Solutions Manual for Advanced Mechanics of Materials and Applied Elasticity, 6th edition By Ansel Ug by ACADEMIAMILL

CHAPTER 1 SOLUTION (1.1) We have

A = 50 × 75 = 3.75(10−3 ) m 2 , θ = 50o , and σ x = P A .

Equations (1.11), with θ = 50 : o

σ x ' = 700(10 3 ) = σ x cos 2 50o = 0.413σ x = 110.18P P = 6.35 kN

or and

3 ) σ x sin 50o = cos 50o 0.492 τ x ' y ' 560(10 σ x 131.2 P = = =

Solving

P = 4.27 kN = Pall

______________________________________________________________________________________ SOLUTION (1.2) Normal stress is

σ x=

125(103 ) 0.05×0.05

= 50 MPa

P A

( a ) Equations (1.11), with θ = 20 : o

cos 2 20o 44.15 MPa = σ x ' 50 =

τ x' y' = −50sin 20o cos 20o = −16.08 MPa = σ y ' 50 cos 2 (20o += 90o ) 5.849 MPa 5.849 MPa y’

44.15 MPa

16.08 MPa

x’ 20 o x

( b ) Equations (1.11), with θ = 45 : o

= σ x ' 50 = cos 2 45o 25 MPa

τ x' y' = −50sin 45o cos 45o = −25 MPa = σ y ' 50 cos 2 (45o += 90o ) 25 MPa 25 MPa y’ 25 MPa

25 MPa x’ 45 o x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.3) From Eq. (1.11a),

σ x = cosσ θ = cos−7530 = −100 MPa x' 2

For θ = 50 , Eqs. (1.11) give then o

σ x' = −100 cos 2 50o = −41.32 MPa

τ x ' y ' = −( −100) sin 50o cos 50o = 49.24 MPa o Similarly, for θ = 140 : σ x' = −100 cos 2 140o = −58.68 MPa τ x ' y ' = −49.24 MPa

58.68 MPa

41.32 MPa 50 o

49.24 MPa

______________________________________________________________________________________ SOLUTION (1.4) Refer to Fig. 1.6c. Equations (1.11) by substituting the double angle-trigonometric relations, or Eqs. (1.18) with σ y = 0 and τ xy = 0 , become or

σ x ' = 12 σ x + 12 σ x cos 2θ

and

τ x ' y ' = 12 σ x sin 2θ

20 = 2PA (1 + cos 2θ )

and

10 = 2PA sin 2θ

The foregoing lead to

2 sin 2θ − cos 2θ = 1

(a)

By introducing trigonometric identities, Eq. (a) becomes

4 sin θ cos θ − 2 cos 2 θ = 0 or tan θ = 1 2 . Hence

Thus,

θ = 26.56o

= 20

gives

P 2(1300)

(1 + 0.6)

P = 32.5 kN

It can be shown that use of Mohr’s circle yields readily the same result. ______________________________________________________________________________________ SOLUTION (1.5) Equations (1.12):

P −150(103 ) = = −76.4 MPa π A 2 (50) 4 P τ max = = 38.2 MPa 2A

σ1 =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.6) Shaded transverse area:

= A 2= at 2(10)(75) = 1.5(103 ) mm 2 Metal is capable of supporting the load

P =σ A =90(106 )(1.5 ×10−3 ) =135 kN

Apply Eqs. (1.11):

P (cos 2 55o ) , P = 114 kN −3 1.5(10 ) P τ x ' y ' = 12(106 ) = − sin 55o cos 55o , P = 38.3 kN −3 1.5(10 )

6 ) σ x ' 25(10 = =

Thus,

Pall = 38.3 kN

______________________________________________________________________________________ SOLUTION (1.7) Use Eqs. (1.11):

P (cos 2 40o ) , P = 51.1 kN −3 1.5(10 ) P sin 40o cos 40o , P = 24.4 kN τ x ' y ' = 8(106 ) = − 1.5(10−3 )

6 = σ x ' 20(10 = )

Thus,

Pall = 24.4 kN

______________________________________________________________________________________ SOLUTION (1.8)

A =15 × 30 = 450 mm 2 Apply Eqs. (1.11):

120(103 ) (cos 2 40o ) 156 MPa = 450 ×10−6 120(103 ) sin 40o cos 40o = τ x' y' = − −131 MPa 450 ×10−6

= σ x'

______________________________________________________________________________________ SOLUTION (1.9) We have A = 450(10

−6

) m2 .

Use Eqs. (1.11):

−100(10 ) (cos 2 60o ) = −55.6 MPa −6 450 ×10 −100(103 ) − sin 60o cos 60o = 96.2 MPa τ x' y' = −6 450 ×10

σ x' =

______________________________________________________________________________________ 3 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (1.10)

θ = 40o + 90o = 130o 150(10 ) P σx = − π (0.08 = −31.83 MPa A = − 0.07 ) 3

Equations (1.11):

σ x' = −31.83cos 2 130o = −13.15 MPa

τ x ' y ' = 31.83sin130o cos130o = −15.67 MPa x’

13.15 MPa

130 o

15.67 MPa

x Plane of weld y’

______________________________________________________________________________________ SOLUTION (1.11) Use Eqs. (1.14),

( 2 x ) + ( −2 xy ) + ( x ) + Fx = 0

( − y 2 ) + ( −2 yz + x ) + (0) + Fy = 0

( z − 4 xy ) + (0) + ( −2 z ) + Fz = 0 3

Solving, we have (in MN m ):

Fx = −3x + 2 xy

Fy =− x + y 2 + 2 yz

Fz = 4 xy + z (a)

Substituting x=-0.01 m, y=0.03 m, and z=0.06 m, Eqs. (a) yield the following values

= Fx 29.4 = Fy 14.5 = Fz 58.8 kN m3 kN m3 kN m3 Resultant body force is thus

Fx2 + Fy2 + Fz2 = 67.32 kN m3

______________________________________________________________________________________ SOLUTION (1.12) Equations (1.14):

− 2c1 y − 2c1 y + 0 + 0 = 0, 0 + c3 z + 0 + 0 = 0, 0+0+0+0=0

4c1 y ≠ 0 c3 z ≠ 0

No. Eqs. (1.14) are not satisfied. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.13) ( a ) No. Eqs. (1.14) are not satisfied. ( b ) Yes. Eqs. (1.14) are satisfied. ______________________________________________________________________________________ SOLUTION (1.14) Eqs. (1.14) for the given stress field yield:

Fx = Fy = Fz = 0

______________________________________________________________________________________ SOLUTION (1.15) y

τ x ' y ' ∆A σ x ' ∆A

y’ 40 ∆A cos20o

50 ∆A cos20o

20o

x’

x 50 ∆A sin20o 60 ∆A sin20o

= ∑ Fx ' 0 :

σ x ' ∆A + 40 cos 2 20o − 60∆A sin 2 20o

−2(50∆A sin 20o cos 20o ) = 0 σ x' = −35.32 + 7.02 + 32.14 = 3.8 MPa F 0: τ ∑= y'

x' y'

∆A − 40∆A sin 20o cos 20o

−60∆A sin 20o cos 20o − 50∆A cos 2 20o +50∆A sin 2 20o = 0 τ x ' y ' = 12.86 + 19.28 + 44.15 − 5.85 = 70.4 MPa ______________________________________________________________________________________ SOLUTION (1.16) 50 ∆A cos25o y’

15 ∆A cos25o 15 ∆A sin25o

25o

90 ∆A sin25o

σ x ' ∆A x’

F 0 : σ ∆A + 50∆A cos 25 ∑= x'

τ x ' y ' ∆A

−90∆A sin 2 25o − 2(15∆A sin 25o cos 25o ) = 0 σ x' = −41.07 + 16.07 + 11.49 = −13.5 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.16 (CONT.)

F 0: τ ∑= y'

x' y'

∆A − 50∆A sin 25o cos 25o

0 −90∆A sin 25o cos 25o − 15∆A cos 2 25o +15∆A sin 2 25o = τ x ' y ' = 19.15 + 34.47 + 12.32 − 2.68 = 63.3 MPa ______________________________________________________________________________________ SOLUTION (1.17) y

τ x' y'

y’ 50 MPa 60 MPa

σ x' 20

x’ x

θ =20o

80 MPa

1 1 (−40 + 60) + (−40 − 60) cos 40o + 50sin 40o 2 2 = 10 − 38.3 + 32.1 = 3.8 MPa 1 τ x' y' = − (−40 − 60) sin 40o + 50 cos 40o 2 = 32.14 + 38.3 = 70.4 MPa

σ x' =

______________________________________________________________________________________ SOLUTION (1.18)

σ x'

τ x' y'

x’

θ y’

90 MPa 25

15 MPa

θ =115o

50 MPa

1 1 (90 − 50) + (90 + 50) cos 230o − 15sin 230o 2 2 = 20 − 45 + 11.5 = −13.5 MPa 1 τ x' y' = − (90 + 50) sin 230o − 15cos 230o 2 = 53.62 + 9.64 = 63.3 MPa

σ x '=

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.19) Transform from θ = 40 to θ = 0 . For convenience in computations, Let o

σx = −160 MPa, σy = −80 MPa, τ xy = 40 MPa and θ = −40o

Then

1 1 (σ x + σ y ) + (σ x − σ y ) cos 2θ + τ xy sin 2θ 2 2 1 1 = (−160 − 80) + (−160 + 80) cos(−80o ) + 40sin(−80o ) 2 2 = −166.3 MPa

σ x'=

1 2 1 = − (−160 + 80) sin(−80o ) + 40 cos(−80o ) 2 = −32.4 MPa

τ x' y' = − (σ x − σ y ) sin 2θ + τ xy cos 2θ

σ y' = σ x + σ y − σ x' = −160 − 80 + 166.3 = −73.7 MPa

For θ = 0 : o

73.7 MPa 32.4 MPa 166.3 MPa x

______________________________________________________________________________________ SOLUTION (1.20) −1 4 53.1o = θ tan = 3 45 + 90 45 − 90 cos106.2o = + σ x' 2 2 = 67.5 + 6.28 = 73.8 MPa 45 − 90 sin106.2o = 21.6 MPa − τ x' y' = 2

21.6 MPa y’

73.8 MPa x’

53.1o

x ______________________________________________________________________________________ 7 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (1.21)

= τ xy 0= θ 70o (a)

σ − 60 sin140o τ x' y' = −30 = −

(b)

σ= 80 = x'

σ = 153.3 MPa

σ + 60 σ − 60 2

cos140o

σ = 231 MPa

______________________________________________________________________________________ SOLUTION (1.22) Equations(1.18) with θ = 60 , σ x = 110 MPa , σ y = 0 , τ xy = 50 MPa give o

o o 1 1 σ x' = 70.8 MPa 2 (110) + 2 (110) cos120 + 50sin120 =

τ x' y' = − 12 (110) sin120o + 50 cos120o = −72.6 MPa o o 1 1 σ y' = 39.2 MPa 2 (110) − 2 (110) cos120 − 50sin120 =

______________________________________________________________________________________ SOLUTION (1.23) Equations(1.18) with θ = 30 , σ x = 110 MPa , σ y = 0 , τ xy = 50 MPa result in o

o o 1 σ x' = 125.8 MPa 2 (110) + 55cos 60 + 50sin 60 =

τ x' y' = − 12 (110) sin 60o + 50 cos 60o = −22.6 MPa o o 1 σ y' = −15.8 MPa 2 (110) − 55cos 60 − 50sin 60 =

______________________________________________________________________________________ SOLUTION (1.24) We have

θ = 25 + 90 = 115o σ x = −10 MPa σ y = 30 MPa

τ xy = 0 (a)

x’

σ x'

τ x' y'

10 MPa 25

y’

30 MPa

1 1 σ x ' = (σ x + σ y ) + (σ x − σ y ) cos 2θ 2 2 1 1 = (−10 + 30) + (−10 − 30) cos 230o = 22.86 MPa 2 2

Thus,

σ= σ= 22.86 MPa w x'

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.24 (CONT.) (b)

1 2 1 = − (−10 − 30) sin 230o = −15.32 MPa 2

τ x' y' = − (σ x − σ y ) sin 2θ

τw

τ w = τ x ' y ' = −15.32 MPa

______________________________________________________________________________________ SOLUTION (1.25) (a)

(b)

0 + 50 2 τ = 49 MPa

80 =+ ( σ1 =

τ= max

(

0 − 50 2 ) +τ 2 2

−50 2 2 = ) + 49 55 MPa 2

50 = 25 MPa 2 0 − 50 2θ s =tan −1[− ] =27 o 2(49) 50 τ x ' y ' = sin 27o + 49 cos 27o = 55 MPa 2

σ='

Thus,

θ s ' = 13.5o y’

25 MPa 25 MPa x’ 13.5o x

55 MPa

______________________________________________________________________________________ SOLUTION (1.26) 80 MPa 40 MPa

θ =-30o x (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.26 (CONT.)

40 + 80 40 − 80 + cos(−60o ) = 60 − 10 = 50 MPa 2 2 σ y = 60 + 10 = 70 MPa

σx =

40 − 80 sin(−60o ) = −17.32 MPa 2

− τ xy =

70 MPa

70 MPa 50 MPa

50 MPa

17.32 MPa

20 MPa

2.68 MPa

50 + 70 50 − 70 2 2 ± ( ) + 2.68= 60 ± 10.35 2 2 = σ 1 70.35 = MPa σ 2 49.65 MPa 2(2.68) 2θ p = tan −1[ ] = −15o 50 − 70

σ 1,2 =

50 − 70 cos(−15o ) + 2.68sin(−15o ) 2 =60 − 9.66 − 0.694 =49.65 MPa

σ x ' =60 + Thus,

θ p " = −7.5o

x 7.5o 49.65 MPa

70.35 MPa ______________________________________________________________________________________ SOLUTION (1.27) τ (MPa)

60 MPa

40 MPa

(60, 50)

50 MPa O α

R C

σ (MPa)

(-40, -50) 40o (CONT.) x’ ______________________________________________________________________________________ 10 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.27 (CONT.) −1 50 45o α tan = = 50 1 2 R =(50 + 502 ) 2 =70.7

85o (70.7) 70.4 MPa τ x ' y ' sin = = 10 − cos85o (70.7) = 3.84 MPa σ x' = ______________________________________________________________________________________ SOLUTION (1.28) τ (MPa)

50 MPa

15 MPa 90 MPa

σ’=20

(-50, -15) x

62.1o

y’ (90, 15) α σ (MPa)

(σx’, τ x’y’) x’

−1 15 12.1o = α tan = 70 1 R =(152 + 702 ) 2 =71.6

13.5 MPa

y’

= = τ x ' y ' 71.6sin 62.1o 63.3 MPa

63.3 MPa 90 MPa

25o

σ x' = −71.6 cos 62.1o + 20

15 MPa

x’

= −13.5 MPa

50 MPa

______________________________________________________________________________________ SOLUTION (1.29) τ (MPa)

σ ' =67.5

x’ 45 MPa

R =22.5 R

53.1

90 MPa

C 106.2o

90 σ (MPa)

73.8o x’ (σx’, −τ x’ y’ )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.29 (CONT.)

= τ x ' y ' 22.5sin = 73.8o 21.6 MPa

σ x' = 67.5 + 22.5cos 73.8o = 73.8 MPa

Sketch of results is as shown in solution of Prob. 1.20. ______________________________________________________________________________________ SOLUTION (1.30) τ (MPa) (a) x’ 1 60 MPa 140o

x’ 70

= R σ

(σ − 60)

σ (MPa)

σ − 60 −30 = τ x' y' = sin(−40o );

σ= 153.3 MPa

(b)

σ x ' =80 =60 +

σ − 60

σ = 231 MPa

[1 − cos(−40o )]

______________________________________________________________________________________ SOLUTION (1.31) ( a ) From Mohr’s circle, Fig. (a):

σ1 = 121 MPa σ2 = −71 MPa τ max = 96 MPa θ p ' = −19.3o θ s ' = 25.7 o τ (MPa) τmax

A(100,60) σ2

2θ p '

σ (MPa)

σ1

B Figure (a) By applying Eq. (1.20):

1 2

50  22,500 + 3600  =± σ 1,2 = 25 96 2 ± 4

σ 1 = 121 MPa

Using Eq. (1.19):

σ 2 = −71 MPa

tan 2θ p = − 12 15 = −0.8

θ p ' = −19.3o

θ s ' = 25.7 o

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.31 (CONT.) ( b ) From Mohr’s circle, Fig. (b):

σ1 = 200 MPa σ2 = τ max = 125 MPa −50 MPa

θ p ' = 26.55o

θ s ' = 71.55o

τ (MPa)

σ2

τmax

2θp’

σ (MPa)

σ1

A(150,-100) Figure (b) Through the use of Eq. (1.20),

σ 1, 2 = 75 ± [22,4500 + 10,000] = 75 ± 125 1 2

σ 1 = 200 MPa σ 2 = −50 MPa Using Eq. (1.19), tan 2θ p = 4 3 :

o = θ p ' 26.57 = θ s ' 71.57o

______________________________________________________________________________________ SOLUTION (1.32) Referring to Mohr’s circle, Fig. 1.15:

σ x ' = σ +2σ + σ −2σ cos 2θ 1

(a)

σ y ' = σ +2σ − σ −2σ cos 2θ 1

τ x ' y ' = σ −2σ sin 2θ 1

(b)

From Eqs. (a),

σ x' + σ y' = σ 1 + σ 2 2θ + sin 2 2θ = 1 , and Eqs. (a) and (b), we have σ x ' ⋅ σ y ' − τ x2' y ' = σ 1 ⋅ σ 2 = const .

By using cos

______________________________________________________________________________________ SOLUTION (1.33) We have

2τ

tan 2θ p = σ x −xyσ y = 502(−(−−70) 190) = −0.583

2θ p = −30.24o

and θ p = −15.12

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.33 (CONT.) Equations (1.18):

σ x ' = 50−2190 + 50+2190 cos( −30.26o ) − 70 sin( −30.26o ) = −70 + 103.65 + 35.275 = σ1 68.93 MPa = σ y' = σ x + σ y − σ x' = −208.9 MPa = σ2 208.9 MPa 15.13o 68.92 MPa

______________________________________________________________________________________ SOLUTION (1.34)

τ max = (

σ x −σ y 2

) + τ xy2

Substituting the given values

140 2 = ( 60+2100 ) + τ xy2 2

τ xy ,max = 114.89 MPa

______________________________________________________________________________________ SOLUTION (1.35) Transform from θ = 60 to θ = 0 with σ x ' = −20 MPa, σ y ' = 60 MPa , o

τ x ' y ' = −22 MPa , and θ = −60o . Use Eqs. (1.18):

σ x = −202+ 60 + −202−60 cos 2(−60o ) − 22sin 2(−60o ) = 59 MPa σy = σ x' + σ y' − σ x = −19 MPa τ xy = −23.6 MPa y

19 MPa 23.6 MPa 59 MPa x

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.36) σy

σycos60o

14 MPa τxy

τxycos60o σxsin60o

30 MPa Area =1

30 MPa

τxysin60o Figure (b)

Figure (a) ( a ) Figure (a):

= σ y 14sin = 60o 12.12 MPa

= = τ xy 14 cos 60o 7 MPa Figure (b):

∑ F = 12.12 cos 60 − τ sin 60 = 0 o

τ xy = 7 MPa (as before)

∑ F = −σ sin 60 + 30 + 7 cos 60 = 0 x

σ x = 38.68 MPa

( b ) Equation (1.20) is therefore:

σ 1, 2 = 38.68+212.12 ± [( 38.682−12.12 ) 2 + 7 2 ]

1 2

or = σ 1 40.41 = σ 2 10.39 MPa MPa, Also,

θ p = 12 tan −1 38.682 (−712) .12 = 13.9 o

Note: Eq. (1.18a) gives, σ x ' = 40.41 MPa . Thus,

θ p ' = 13.9 o 10.39 MPa 40.41 MPa x’ θp’

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.37) σy τxy Figure (a)

σx

Figure (a):

= σ x 100 = cos 45o 70.7 MPa

45o 70.7 MPa = σ y 100sin = = τ xy 100 = cos 45o 70.7 MPa Now, Eqs. (1.18) give (Fig. b):

o 70.7 + 0 + 70.7 sin 240= 9.47 MPa σ= x'

τ x ' y ' =−0 + 70.7 cos 240o =−35.35 MPa o σ= 70.7 − 0 − 70.7 sin 240= 131.9 MPa y'

σy

σx’ τxy

x’ σx y’

τx’y’

Figure (b)

n 30o

σy’

______________________________________________________________________________________ SOLUTION (1.38)

−70sin 30o = −35 MPa σy =

cos 30o 60.6 MPa = τ xy 70 = ( a ) Figure (a):

∑ F = −150 + 0.5σ + 60.6(0.866) = 0 x

σ x = 195 MPa

35cos30o 150 MPa

60.6cos30o

30o Area=1

60.6sin30o σxsin30

Figure (a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.38 (CONT.) ( b ) Equation (1.20): 195 −35 σ 1,2 = ± ( 1952+35 ) 2 + 60.62  2

or Also,

σ 1 = 210 MPa

1 2

σ 2 = −50 MPa

2 ( 60.6 ) o θ p = 12 tan −1 195 + 35 = 13.89

Equation (1.18a): Thus,

σ x' = 80 + 115cos 2(13.89o ) + 60.6sin 2(13.89o ) = 210 MPa

θ p ' = 13.89 o

50 MPa 210 MPa x’ θp’

______________________________________________________________________________________ SOLUTION (1.39) For pure shear, σ 1 = −σ 2 : pr t

from which

σ 1 = pr t

= − 2prt + 2πPrt

σ 2 = 2prt − 2πPrt

P = 3πpr 2

______________________________________________________________________________________ SOLUTION (1.40) σθ y Table D.4:

σ +σa

A = 2πrt J = 2πr 3t

τ xy

Figure (a)

Stresses are (Fig. a):

σ=

σ= a

−P A

pr 2t

−30(103 )π 2π (0.12)(0.005) 6

− 25 MPa

4(10 )120 2(5)

= 48 MPa

= σ θ 2= σ a 96 MPa 3

−10π (10 ) τ xy = −JTr = 2π (0.12 = −69.4 MPa )(0.005) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.40 (CONT.) Hence,

σ x = 48 − 25 = 23 MPa

σ y = 96 MPa

Therefore, we have

1 2

τ max = ± ( 23−296 ) 2 + 69.42  = ±78.4 MPa Also

σ '=

and

1 2

y’

(23 + 96) = 59.5 MPa

−96 − 12 tan −1 2(23−69.4) = −13.87o θs =

θs ' '

59.4 MPa

78.4 MPa

Equation (1.18b) with θ s = −13.87 : o

−16.99 − 61.42 = −78.4 MPa τ x' y' = Thus,

x x’

Figure (b)

θ s ' ' = 13.87o

______________________________________________________________________________________ SOLUTION (1.41)

= A 2= π rt 2π (60)(4) = 1508 mm 2 3 J 2π r 3t 2π (60)= (4) 5.429 ×106 mm 4 = =

σ= y

pr t

5(60) 4

= 75 MPa

σy

( 0.05 ) τ xy = − TrJ = − 5600 .429 (10 ) −6

= −5.526 MPa P σ x = 2prt + PA = 37.5 + 1508 ( P in newtons )

σx τ xy

Thus

σ1 =

σ x +σ y 2

σ −σ

+ ( x 2 y ) 2 + τ xy2

Substituting the numerical values gives

[

]

80 = 56.3 + 331.6 × 10 −6 P + ( −18.75 + 331.6 × 10 −6 P ) 2 + ( −5.526) 2 2 Solving,

P = 64.01 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.42) At point A, T = 8 kN and P = 400 kN

P A

− = − σx =

τ xy

4(400 ×103 ) = −50.9 MPa π (0.1) 2

16T 16(8 ×10 ) = = 40.7 MPa τ xy = πd3 π (0.1)3

σx

Hence

= τ max

(

σx 2 2

) + τ xy2

= (−25.45) 2 + (40.7) 2 = 48 = MPa R

σ avg =

τ y

σx

= −25.45 MPa

tan 2θ p " =

40.7 , 25.45

σ 2 2θp"

θ p " = 29o

O σavg

σ1

______________________________________________________________________________________ SOLUTION (1.43)

θ =α + 90 =50 + 90 =140o 120 ×103 = 95.5 MPa π P (0.04) 2 σx = θ 4 A x 16(1.5 ×10−3 ) 16T = = 119.4 MPa τ τ xy = 3 π (0.04)3 πd = σx

y' Equations(1.18):

95.5 2

cos 280o + 119.4sin 280o = + −61.5 MPa σw = σ x' = and

95.5 2

sin 280o − 119.4 cos 280o = 47.02 − 20.73 = 26.3 MPa τw = τ x' y' = − 61.5 MPa

x’

140o x

26.3 MPa

y’ ______________________________________________________________________________________ 19 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (1.44)

A = π4 (1802 − 1202 ) = 14.14 ×103 mm 2 π J = 32 (1804 − 1204 ) = 82.70 ×106 mm 4 700 σ x =− PA =− 14.14 =−49.5 MPa,

τ xy=

Tr J

20(90)

σ y =0

= 21.77 MPa

82.70×10−6

Equation (1.20) is therefore 2 49.5 2 σ max = σ1 = − 49.5 2 + ( 2 ) + (21.76)

= 8.205 MPa ______________________________________________________________________________________ SOLUTION (1.45)

P = τ 0 Lt M = τ 0 Lth A = 2ht I = 121 t ( 2h ) 3 = 23 th 3

τ 0 Lt A

P M

τ L

Axial stress: σ a = PA = 20h

Bending stress: σ b = Mc I = Point A

3τ 0 L 2h

τ0

σ a + σ b = τ2 hL

From Eqs. (1.20) and (1.22), we obtain

σ 1, 2 = τ hL ± τ 0 ( Lh ) 2 + 1 0

τ max = τ 0 ( Lh ) 2 + 1 Point B

σ b − σ a = τ hL

B Hence

σ 1 = τ hL 0

τ max = 2τ h

σ2 = 0

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.46) y T

P M

A= π (302 − 152 )= 2.121(10−3 ) m 2 I = π4 (304 − 154 )= 0.596(10−6 ) m 4

J = 2I

σa +σb

τ xy

We have

σ a=

P A

σ= b

Mr I

50(103 )

= 23.58 MPa

2.12(10−3 )

200(0.03)

= 10.07 MPa

0.596(10−6 )

−500(0.03) τ xy = −JTr = 1.192(10 = −12.58 MPa ) −6

Thus,

σ x = 23.58 + 10.07 = 33.65 MPa

σ ' = 16.83 MPa

τ (MPa)

(33.65, 12.58) 2θp’ σ (MPa)

Figure (a)

16.83

From Mohr’s circle (Fig. a):

12.582 + 16.832 = 21.01 MPa

o .58 θ p ' = 12 tan −1 12 16.83 = 18.39

σ 1 = 16.83 + 21.01 = 37.84 MPa σ 2 = −4.18 MPa Results are shown in Fig. (b). θ p’

x x’

Figure (b)

37.84 MPa 4.18 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.47) A

y 30

20o

τ 0 = τ xy ( a ) At θ = −60

0= or

C τ0

σx’

-60o

y’ Figure (a)

x’

(Fig. a): σ x +σ y 2

σ x −σ y 2

cos 2( −60o ) + τ 0 sin 2( −60o )

0 = 0.5σ x + 1.5σ y − 1.732τ 0

We also have

(a)

σ x −σ y

τ0 = − 2 sin 2(−60o ) + τ o cos 2(−60o ) or

σ x = 3.464τ 0 + σ y

(b)

Substituting Eq. (b) into (a), we obtain σ y = 0 . Results are shown in Fig. b. y

τ xy = τ 0

σ x = 3.464τ 0

Figure (b)

x Alternatively, using an element ABC (Fig. c):

σx

τ0

τ0 τ0

30o

Figure (c)

Area=1

∑ F = 0.5σ − 0.866τ − 0.866τ = 0 x

or σ x = 3.464τ 0 , as before. Stresses on planes at 20 , taking θ = −70 (Fig. b): o

o o 3.464 σ 20 = [ 3.464 2 + 2 cos( −140 ) + sin( −140 )]τ 0 = −0.237τ 0 o

o o τ 20 = [ − 3.464 2 sin( −140 ) + cos( −140 )]τ 0 = 0.347τ 0 o

(CONT.) ______________________________________________________________________________________ 22 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.47 (CONT.) ( b ) Principal stresses: 3.464τ 3.464τ σ 1,2 = ± ( 2 ) 2 + τ 02  2 o

σ 1 = 3.732τ 0

1 2

σ 2 = −0.268τ 0

The maximum principal stress is on plane inclined at τ o θ p ' = 12 tan −1 1.732 τ = 15 0

______________________________________________________________________________________ SOLUTION (1.48) At a critical point on the shaft surface, the state of stress of stress is as shown in Fig. (a). y

σ1

τ max

σx

θp''

τ xy

Figure (a) We have

σ2

Figure (b)

σ x = − PA − MrI 3

) )(0.075) = − π81(10 − 13(10 = −43.818 MPa (0.075)2 π (0.075)4 4 3

×10 )0.075 τ xy = − TrJ = − (15.6 = −23.54 MPa π (0.075) 2 4

Therefore,

σ 1, 2 = −432.818 ± [( 43.2818 ) 2 + ( −23.54) 2 ]

1 2

and

σ 1 = 10.248 MPa, σ 2 = −54.066 MPa 1 τ max = 2 (σ 1 − σ 2 ) = 32.157 MPa .54 ) θ p ' ' = 12 tan −1 243( 23.818 = 23.53o

Results are shown in Fig. (b). ______________________________________________________________________________________ SOLUTION (1.49) Apply Eqs. (1.20) to Fig. P1.49b, for θ = −30 : o

σ xb = −40sin 2(−30o ) = 20 3 MPa σ yb = −20 3 MPa

(b)

τ xyb = −40 cos 2(−30 ) = −20 MPa o

(CONT.) ______________________________________________________________________________________ 23 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.49 (CONT.) Now apply Eqs. (1.18) to Fig. P1.49c, for θ = −60 : o

σ xc = 10sin 2(−60o ) = −5 3 MPa σ yc = 5 3 MPa

(c)

τ xyc = −5 MPa 10 cos 2(−60 ) = o

Superposing stresses in Eqs. (b) and (c) and those in Fig. P1.49a, we obtain Fig. (a). y

15 3 MPa 15 3 MPa x

45 MPa Figure (a) Referring to Fig. (a):

[

σ 1, 2 = 0 ± (15 3 ) 2 + ( −45) 2 or When

σ 1 = 51.96 MPa

]

1 2

σ 2 = −51.96 MPa

θ p ' = 12 tan −1 22((15−453)) = −30o

is substituted into Eq. (1.18a), we have 51.96 MPa (Fig. b). x

θp' x’

51.96 MPa 51.96 MPa Figure (b) ______________________________________________________________________________________ SOLUTION (1.50) Apply Eqs. (1.18) to Fig. P1.50a, for θ = −15 , to obtain stresses in Fig. (a): o

σ xa = − 302 − 302 cos 2(−15o ) = −27.99 MPa

σ ya = −15 + 15cos 2(−15o ) = −2.01 MPa

τ xya = 15sin 2(−15o ) = −7.5 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.50 (CONT.) Superposition of stresses in Figs. (a) and P1.50b gives Fig. (b). 2.01 MPa

47.99 MPa

27.99 MPa

27.99 MPa 30o

30o 7.5 MPa

7.5 MPa

Figure (b)

Figure (a) Apply Eq. (1.20) to Fig. (b):

σ 1, 2 = −27.992+47.99 ± [14 ( −27.99 − 47.99) 2 + ( −7.5) 2 ]

1 2

or When

σ 1 = 48.72 MPa,

σ 2 = −28.72 MPa

5) o θ p = 12 tan −1 −( 272.(99−7+.47 .99 ) = 5.58

is substituted into Eq. (1.18a), we obtain –28.72 MPa (Fig. c). 48.72 MPa

28.72 MPa 35.58o x

Figure (c) ______________________________________________________________________________________ SOLUTION (1.51) Equations (1.18) are applied to Fig. P1.51a, for θ = −30 : o

σ xa = 20+2 30 + 20−2 30 cos 2(−30o ) = 22.5 MPa

σ ya = 25 − (−5) cos 2(−30o )= 27.5 MPa

τ xya =−(−5) sin 2(−30o ) =−4.33 MPa These stresses and that of Fig. P1.51b are superimposed to yield Fig. (a). y

14.33 MPa 37.5 MPa x 22.5 MPa

(CONT.) Figure (a) ______________________________________________________________________________________

______________________________________________________________________________________ 1.51 (CONT.) Principal stresses are thus

σ 1, 2 = 37.5+2 22.5 ± [( 37.5−2 22.5 ) 2 + 14.332 ]

1 2

= σ 1 46.17 = σ 2 13.83 MPa MPa Hence

We have

τ max =

1 2

(σ 1 − σ 2 ) = 16.17 MPa

θ p = 12 tan −1 237( −.514−22.33.5) = −31.2 o

Equation (1.18a) results in

σ x ' = 37.5+2 22.5 + 37.5−2 22.5 cos(−62.4o ) − 14.33sin(−62.4o ) = 46.17 MPa

Therefore

θ p ' = 31.2 o

Results are shown in a properly oriented element in Fig. (b). 13.83 MPa

θp'

Figure (b)

x’ 46.17 MPa

______________________________________________________________________________________ SOLUTION (1.52) State of stress is represented by Mohr’s circle in Fig. (a). From this circle, we determine

σ x = −40 MPa σ y = 20 MPa θ p ' = tan 1 2

−1 4 3

= 26.57

-60

(-σx,-40)

τ (MPa) (σy,40) O

2θ

σ (MPa)

Figure (a)

Results are shown in Fig. (b). 50 MPa

θp''

Figure (b)

60 MPa

40 MPa ______________________________________________________________________________________ 26 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (1.53) 27 MPa

y’

x’

21 MPa

θ x

σ x + σ y = σ x' + σ y' 45 − 30 = −27 + σ ; σ= 42 MPa 45 − 30 45 + 30 + 42 = cos 2θ + 15sin 2θ σ x' = 2 2

= 34.5 37.5cos 2θ + 15sin 2θ τ x' y' = −21 = −37.5sin 2θ + 15cos 2θ

(1)

Multiply this by − 2.5 :

= 52.5 93.75sin 2θ − 37.5cos 2θ

(2)

Add Eqs. (1) and (2),

= = 87 108.75sin 2θ , 2θ 53.13o or

θ = 26.6o

______________________________________________________________________________________ SOLUTION (1.54) State of stress is represented by Mohr’s circle in Fig. (a). τ (MPa)

(100,τxy) 2θp’ 60

110

σ (MPa)

Figure (a)

(20,-τxy) Referring to this circle, we obtain the results (Fig. b).

θ p ' = 12 tan −1 43 = 18.43o 20 MPa

50 MPa

10 MPa

100 MPa

x x’

θp'

110 MPa

30 MPa (b)

(a)

Figure (b) ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.55)

σx = −18 MPa τ xy = −15 MPa σ x ' = 60 MPa σ y = 30 MPa From Equation (1.18a): or Solving

30 = 602−18 + 602+18 cos 2θ1 − 15 sin 2θ1 13 cos 2θ1 − 5 sin 2θ1 − 3 = 0 2θ1 = 56.52 o

θ1 = 28.26o

We have

σ y ' = σ x + σ y − σ x ' = 12 MPa

Equation (1.18b) gives

τ x' y' = − 602+18 sin 56.52o − 15cos 56.52o = −40.8 MPa

______________________________________________________________________________________ SOLUTION (1.56) We have

= σ

4M π r3

4(21π )103

= 84 MPa

π (0.1)3

State of stress is represented by Mohr’s circle in Fig. (a). τ (MPa)

56 (84,τ)

τ C

84 MPa

σ (MPa)

Figure (a) 1 2

τ =(56 − 42 ) =37.04 MPa 2

Thus Hence

τJ r

(37.04)(106 )π (0.13 ) 2

= 58.18 kN ⋅ m

P 2= = π fT 2π (20)58.18 = 7311 kW

______________________________________________________________________________________ SOLUTION (1.57) τ σ1 τxy

60o σ1=2σ2

σ2

σ1

r = 2σ 22−σ 2 = σ22

(CONT.) ______________________________________________________________________________________ 28 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.57 (CONT.) From Mohr’s circle,

τ x ' y ' = ( σ2 ) sin 60o = 0.433σ 2 2

Therefore = 30 We have Solving

σ= 2

0.433 = σ2 σ 2 69.28 MPa

pr 2t

= 69.28

p ( 2250 ) 69.28 ×5=

p = 2.771 MPa

______________________________________________________________________________________ SOLUTION (1.58) pr t

A pr 2t

+ 2πPrt

Mohr’s circle representing stress at point A is shown in Fig. (a). τ (MPa) (84,τxy) 60o O

σ (MPa)

(56,-τxy) Figure (a) From this circle: Then gives

.45 ) = 90 p 42 = p0(.0005

3 98(10 = )

467(0.45) 2(0.005)

p = 467 kPa

P + 2π (0.45)(0.005)

P = 1088 kN

______________________________________________________________________________________ SOLUTION (1.59) σ’=70 τ (MPa)

y’

(40,-τxy)

(100,τxy)

35o 70o

σ (MPa) x’

= r

= 36.62 MPa

30 cos35o

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.59 (CONT.) (a)

τ xy = −36.62sin 35o = −21 MPa

( b ) Because of symmetry:

τ x' y' = −τ xy = 21 MPa

and

σ x + σ y = σ x ' + σ y ' = 140 MPa

gives

σ y ' = 40 MPa

______________________________________________________________________________________ SOLUTION (1.60) State of stress is represented by Mohr’s circle in Fig. (a). τ (MPa) (-12,20) -14

(σx,-20)

σ (MPa)

Figure (a) ( a ) Using this circle, we write

[

τ max = ( σ 2+12 ) 2 + 20 2

and Solving,

]

1 2

τ max = 14 + OC = 14 + 12 (σ x − 12)

σ x = 186 MPa

Note that, alternately,

[

]

− 14 = σ x 2−12 − ( σ x 2+12 ) 2 + 20 2 2 yields σ x = 186 MPa , as before. ( b ) We have

σ 1, 2 = 1862−12 ± [( 1862+12 ) 2 + 20 2 ]

1 2

or and Also

σ 1 = 188 MPa τ max =

1 2

σ 2 = −14 MPa

(σ 1 − σ 2 ) = 101 MPa

2 ( 20 ) o θ p ' = 12 tan −1 186 +12 = 5.71

(CONT.) ______________________________________________________________________________________ 30 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.60 (CONT.) Results are shown in Fig. (b). 14 MPa 101 MPa

188 MPa x

θp'

Figure (b) ______________________________________________________________________________________ SOLUTION (1.61) ( a ) σ 1 96.05 = = MPa σ 2 23.95 = MPa σ3 0 (b)

(τ 12 ) max = (τ 13 ) max = (τ 23 ) max =

1 2 1 2 1 2

(σ 1 − σ 2 ) = 36.05 MPa (σ 1 − σ 3 ) = 48.03 MPa (σ 2 − σ 3 ) = 11.98 MPa

Plane of (τ 12 ) max is shown in Fig. (a). Other maximum shear planes are sketched similarly. σ2 (τ12)max

σ1 σ3

Figure (a)

28.15o x

______________________________________________________________________________________ SOLUTION (1.62)

A 2= π rt 2π (60)(4) = = 1508 mm 2 pr 2(60) σ= y t= 4 = 30 MPa 3

5(10 ) σ x =− PA + 2prt =− 1508(10 + 15 = 11.684 MPa ) −6

τ (MPa) y

σ’ σx’

30 MPa

11.684 MPa

64o 11.684

σ (MPa)

30 (CONT.) ______________________________________________________________________________________ 31 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.62 (CONT.) 1 σ '= 20.84 MPa 2 (30 + 11.684) = 1 r= 9.158 MPa 2 (30 − 11.684) =

(a)

σ x' = σ '− r cos 64o = 16.82 MPa

( b= ) τ x' y'

r= sin 64o 8.231 MPa

______________________________________________________________________________________ SOLUTION (1.63) 0.009π MN

σ 1 = prt π σ 2 = 2prt − 0.2009 πrt

x’

60o

y’

50p Figure (a)

25p-2.5

Equilibrium of x ' and y ' directed forces results in (Fig. a): or and

21 − 50 p( 23 ) 2 − ( 25 p − 2.5)( 12 ) 2 = 0 pall = 494 kPa 7 + ( 25 p − 2.5)( 43 ) − 50 p( 43 ) = 0

from which

p = 547 kPa

______________________________________________________________________________________ SOLUTION (1.64) Direction cosines are:

l1 = 3 2 m1 = 1 2 n1 = 0 l 2 = − 1 2 m2 = 3 2 n 2 = 0 l3 = 0 m3 = 0 n3 = 1 Equation (1.28a) is thus

σ= 20( 34 ) + 0 + 0 + 2(12)( 23 )( 12 ) + 0 += 0 25.392 MPa x' Similarly, applying Eqs. (1.28b) through (1.28e), we obtain [τ i ' j ' ] : (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.64 (CONT.)

 25.392 −2.66 −7.99   −2.66 −5.392 16.16  MPa    −7.99 16.16 6  Then, Eqs. (1.34) result in

I 2 = I 2 ' = −349 (MPa) 2

I= I= 26 MPa 1 1'

I 3 = I 3 ' = −6464 (MPa)3 ______________________________________________________________________________________ SOLUTION (1.65) Direction cosines are:

l1 = 3 2 m1 = 1 2 n1 = 0 l 2 = − 1 2 m2 = 3 2 n 2 = 0 l3 = 0 m3 = 0 n3 = 1 Equation (1.28a) is therefore

σ x ' = 60( 43 ) + 0 + 0 + 2( 40)( 23 )( 12 ) + 0 + 0 = 20[( 94 ) + 3]= 79.64 MPa Similarly, applying Eqs. (1.28b) through (1.28e), we obtain [τ i ' j ' ] :

 79.64 −5.98 −44.64   −5.98 −19.64 2.68  MPa   −44.64 2.68 20  Then, Eqs. (1.34) lead to

I 2 = I 2 ' = −2400 (MPa) 2

I= I= 80 MPa 1 1'

I= I= 8000 (MPa)3 3 3' ______________________________________________________________________________________ SOLUTION (1.66) Referring to Appendix B: and Thus,

σ 1 = 13.212 MPa l1 = 0.9556

τ max =

1 2

σ 2 = 5.684 MPa

m1 = 0.1688

σ 3 = −8.896 MPa

n1 = 0.2416

(σ 1 − σ 3 ) = 11.054 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.67) Referring to Appendix B: and

σ 1 = 66.016 MPa l1 = 0.9556

σ 2 = 28.418 MPa

m1 = 0.1688

σ 3 = −44.479 MPa

n1 = 0.2416

______________________________________________________________________________________ SOLUTION (1.68) Referring to Appendix B: Thus,

σ 1 = 30.493 MPa

τ max =

1 2

σ 2 = 12.485 MPa

σ 3 = −16.979 MPa

(σ 1 − σ 3 ) = 23.736 MPa

______________________________________________________________________________________ SOLUTION (1.69) Referring to Appendix B:

= σ 1 24.747 = MPa σ 2 8.480 = MPa σ 3 2.773 MPa and

l1 = 0.6467

m1 = 0.3958

n1 = 0.6421

______________________________________________________________________________________ SOLUTION (1.70) ( a ) Equation (1.32) becomes

(30 − σ p ) 0 0 −σ p 20 0

20 0 =0 −σ p

Expanding,

− σ p [σ p (σ p − 30) − 400] = 0

or Thus

σ p = 0,

σ p = −10,

σ 1 = 40 MPa,

σ p = 40

σ 2 = 0,

σ 3 = −10 MPa

( b ) For σ 1 = 40 MPa :

(30 − 40)l1 + (0)m1 + 20n1 = 0, (0)m1 = 0, 2

l1 = 2n1 m1 = 0

The condition l1 + 0 + n1 = 1 gives

( 2n1 ) 2 + n12 = 1, Thus

l1 = 25 ,

m1 = 0,

n1 = 15

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.71) y 60o

y’

x,x’

30o z

z’

60o

Figure (a)

( a ) At point (3,1,5) with respect to xyz axis, we have [τ ij ] :

10  0   0

0 −4

0 0  MPa 8 

Then, Eqs. (1.34) result in

(a)

I 2 = 8 (MPa) 2

I1 = 14 MPa

I 3 = −320 (MPa)3

Direction cosines of x’ y’ z’, referring to Fig. (a) are

l1 = 1 m1 = 0 n1 = 0 l 2 = 0 m2 = 1 2 n2 = 3 2 l 3 = 0 m3 = − 3 2 n 3 = 1 2

Now Eqs. (1.28) and (a) give [τ i ' j ' ] :

0 0  10   5 3 3  MPa 0  0 3 3 −1    Thus, Eqs. (1.34) yield

I 2 ' = 8 (MPa) 2

I1 ' = 14 MPa

as before.

y” x

Figure (b)

26.56o

z,z”

I 3 ' = −320 (MPa)3

x”

( b ) Direction cosines are (Fig. b):

l1 = 2

5 m1 = − 1 5 n1 = 0 l 2 = 1 5 m2 = 2 5 n 2 = 0 l3 = 0 m3 = 0 n3 = 1

(CONT.) ______________________________________________________________________________________ 35 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.71 (CONT.) With these and Eq. (a), Eqs. (1.28) yield [τ i ' j ' ] :

7.2 5.6 0  5.6 −1.2 0  MPa   0 0 8  Thus, Eqs. (1.34) result in

I 2 '' = 8 (MPa) 2

I1 '' = 14 MPa

I 3 '' = −320 (MPa)3

The I’s are thus invariants. ______________________________________________________________________________________ SOLUTION (1.72) Introducing the given data into Eq. (1.28a), we obtain

σ x ' = 12( 12 ) 2 + 10( 23 ) 2 + 14(0) + 2[6( 12 )( 23 )] + 0 + 0 = 15.696 MPa

Remaining stress components are determined in a like manner. The result, [τ i ' j ' ] , is

 15.696 −3.866 7.089   −3.866 6.304 −6.294  MPa    7.089 −6.294 14. 

______________________________________________________________________________________ SOLUTION (1.73) Equations (1.34) become

I1 = σ x + σ y

I 2 = σ x ⋅ σ y − τ xy2

I3 = 0

Equation (1.33) is then

σ 3p − (σ x + σ y )σ p2 + (σ xσ y − τ xy2 )σ p = 0

σ p2 − (σ x + σ y )σ p + (σ xσ y − τ xy2 ) = 0

Solution of this quadratic equation is 1

σ 1, 2 =

σ x +σ y

± 12 [σ x2 + 2σ xσ y + σ y2 − 4(σ x + σ y − τ xy2 )] 2

σ x +σ y

± [( x 2 y ) 2 + τ xy2 ]

σ −σ

1 2

______________________________________________________________________________________ SOLUTION (1.74) Referring to Appendix B, we obtain the following values. (a) and (b)

σ 1 = 12.049 MPa σ 2 = −1.521 MPa σ 3 = −4.528 MPa l1 = 0.6184 m1 = 0.5333 n1 = 0.5772 σ 1 = 19.238 MPa σ 2 = 13.704 MPa σ 3 = 4.648 MPa m1 = 0.3862 l1 = 0.3339 n1 = 0.8599

and ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.75) ( a ) Direction cosines are: y C

l = 4 5 = 0.8 m = 3 5 = 0.6 n=0

σ x’ B

Equation (1.40) is thus

σ = 100(0.8) 2 + 60(0.6) 2 + 2(40)(0.8)(0.6) = 124 MPa

Equations (1.26) yield

px = 100(0.8) + 40(0.6) = 104 MPa p y = 40(0.8) + 60(0.6) = 68 MPa

= pz 80(0.6) = 48 MPa Equation (1.41) is then 1 2

τ = [1042 + 682 + 482 − 1242 ] = 48.66 MPa ( b ) Direction cosines are:

y τ

σ y’ z

B 2

F 4

l=0 m=2

20 = 0.447

n=4

20 = 0.894

Equation (1.40) results in

σ = 60(0.447) 2 + 20(0.894) 2 + 2(80)(0.447)(0.894) = 91.912 MPa

Equations (1.26) yield

= px 40(0.447) = 17.88 MPa p y = 60(0.447) + 80(0.894) = 98.34 MPa pz = 80(0.447) + 20(0.894) = 53.64 MPa

Equation (1.41) leads to 1 2

τ = [17.882 + 98.342 + 53.642 − 91.9122 ] = 66.482 MPa ( c ) Direction cosines are:

l = 0.512

m = 0.384

n = 0.768

Equation (1.40) is therefore

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.75 (CONT.)

σ = 100(0.512) 2 + 60(0.384) 2 + 20(0.768) 2

+2[40(0.512)(0.384) + 80(0.384)(0.768)] = 109.77 MPa

Equations (1.26) yield

px = 100(0.512) + 40(0.384) = 66.56 MPa p y = 40(0.512) + 60(0.384) + 80(0.768) = 104.96 MPa pz = 80(0.384) + 20(0.768) = 46.08 MPa

Equation (1.41) gives 1 2

τ = [66.562 + 104.962 + 46.082 − 109.77 2 ] = 74.3 MPa ______________________________________________________________________________________ SOLUTION (1.76) (a)

Direction cosines are: y σ

x’

C 2

13 = 0.555

m = 3 13 = 0.882 n=0

l=2 x

Equation (1.40) is then

σ = 100(0.555) 2 + 60(0.832) 2 + 2(40)(0.555)(0.832) = 109.277 MPa

Equations (1.26) lead to

px = 100(0.555) + 40(0.832) = 88.78 MPa p y = 40(0.555) + 60(0.832) = 72.12 MPa

= pz 80(0.832) = 66.56 MPa Equation (1.41) gives then 1 2

τ = [88.782 + 72.122 + 66.562 − 109.277 2 ] = 74.65 MPa ( b ) Direction cosines are:

y σ y’

B 1

G z’

l=0 m =1

5 = 0.447

n=2

5 = 0.894

Thus, the results are the same as those obtained in Solution of Prob. 1.75b.

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.76 (CONT.)

 



 



( c ) We have rg = 3i , re = 2 j , ra = k . Equation (P1.75) is therefore

 x − 3 y z  − 3 2 0 = 0    − 3 0 1

2 x + 3 y + 6z = 6

Direction cosines are:

2 2 2 + 32 + 62

= 72

m = 73

n = 76

With these and given stresses, Eqs. (1.40) and (1.26) yield

σ = 102.449 MPa

and

px = 45.714 MPa

p y = 105.714 MPa

pz = 51.429 MPa

Substituting the above values into Eq. (1.41), we obtain

τ = 73.582 MPa

______________________________________________________________________________________ SOLUTION (1.77) See: Hint, Prob. 1.70:

2 32 +12 + 2 2

= 214

m = 114

n = − 314

Equation (1.40) gives

σ = 20( 214 ) 2 + 30( 114 ) 2 + 50( − 314 ) 2 3 2 2 +2[10( 14 )( 114 ) − 10( 14 )(− 14 )] = 51.43 MPa

Equation (1.41):

τ = {[20( 214 ) + 10( 114 ) − 10( − 314 )]2 1

3 2 2 +[10( 14 ) + 30( 114 ) + 0]2 + [−10( 14 ) + 0 + 50(− 14 )]2 − 51.432 }2

= 7.413 MPa

______________________________________________________________________________________ SOLUTION (1.78) Direction cosines are

l = cos 35o = 0.8192

m = cos 60o = 0.5

n = cos 73.6o = 0.2823

Equation (1.40) results in

σ = 60(0.8192) 2 − 40(0.5) 2 + 30(0.2823) 2

+ 2[20(0.8192)(0.5) − 5(0.5)(0.2823) + 10(0.8192)(0.2823)] = 52.25 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.78 (CONT.) Equations (1.26):

60(0.8192) + 20(0.5) + 10(0.2823) p= = 61.975 MPa x py = 20(0.8192) − 40(0.5) − 5(0.2823) = −5.0275 MPa

= pz 10(0.8192) − 5(0.5) + 30(0.2823) = 14.161 MPa Equation (1.41) is thus 1

= τ [(61.975) 2 + (−5.0275) 2 + (14.161) 2 − (52.25) 2 ] 2 = 36.56 MPa ______________________________________________________________________________________ SOLUTION (1.79) Direction cosines are

l = cos 40o = 0.766

m = cos 75o = 0.259

n = cos 54 o = 0.588

Equation (1.40):

σ = 40(0.766) 2 + 20(0.259) 2 + 20(0.588) 2

+ 2[40(0.766)(0.259) + 0 + 30(0.766)(0.588)] = 23.47 + 1.34 + 6.91 + 42.9 = 74.62 MPa

Equation (1.41) gives

τ = {[40(0.766) + 40(0.259) + 30(0.588)]2 1

+ [40(0.766) + 20(0.259) + 0]2 + [30(0.766) + 0 + 20(0.588)]2 − 74.62 2 } 2 1

= [3438.6 + 1283.1 + 1206.9 − 5568.1] 2 = 18.99 MPa ______________________________________________________________________________________ SOLUTION (1.80) Note: Planes of maximum shear stresses can be determined upon following a procedure similar to that used in Solution of Prob. 1.61. ( a ) From Problem 1.74a: Thus,

−1.521 MPa −4.528 MPa σ1 = 12.049 MPa σ2 = σ3 = (τ 13 ) max = (τ 12 ) max = (τ 23 ) max =

1 2 1 2 1 2

(σ 1 − σ 3 ) = 8.289 MPa (σ 1 − σ 2 ) = 6.785 MPa (σ 2 − σ 3 ) = 1.503 MPa

( b ) From Problem 1.74b:

= σ 1 19.238 = MPa σ 2 13.704 = MPa σ 3 4.648 MPa (CONT.) ______________________________________________________________________________________ 40 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 1.80 (CONT.)

(τ 13 ) max = (τ 12 ) max =

Thus,

1 2 1 2 1 2

(τ 23 ) max =

(σ 1 − σ 3 ) = 7.295 MPa (σ 1 − σ 2 ) = 2.767 MPa (σ 2 − σ 3 ) = 4.528 MPa

______________________________________________________________________________________ SOLUTION (1.81) We have

σ 1 = 48 MPa

σ 2 = 36 MPa

σ 3 = −72 MPa

( a ) From Eqs. (1.43) and (1.44): 1 2

2 2 2 1 ] 53.96 MPa τ= oct 3 [(48 − 36) + (36 + 72) + ( −72 − 48)= 1 σ oct= 3 (48 + 36 − 72)= 4 MPa

( b ) Using Eq. (1.45),

τ max =

1 2

(48 − 36) = 6 MPa

______________________________________________________________________________________ SOLUTION (1.82) (a)

Using Eq.(1.32), characteristic stress determinate, is

50 − σ p 40 0

40 0 −30 = 0 0 0 20 − σ p

Equations (1.34) give

I1 = 50 − 30 + 20 = 40 MPa I 2 =50(−30) + 50(20) + (−30)(20) − 402 =−2700 (MPa) 2

I 3= 50(−30)20 + 2(40)(0)(0) − 50(0) 2 − (−30) 2 (0) 2 − 20(40) 2

= −62, 000 (MPa)3 Equation (1.33) becomes

σ 3p − I1σ p2 + I 2σ p + 62, 000 = 0

Solving, σ 1 = 66.568 MPa,

σ 2 = 20 MPa, σ 3 = −46.568 MPa Note that σ 1 + σ 2 + σ 3 = σ x + σ y + σ z = 40 MPa. Equation (B.6)

result in

= l1 0.924, = m1 0.383, = n1 0 2

Thus, l1 + m1 + n1 = 1 OK

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.82 (CONT.) (b)

Equation (1.45), substituting the values obtained for principle stresses, leads to

(c)

Equations (1.44) and (1.43) are give

τ 13 = τ 23 = τ 12 =

1 2 1 2 1 2

(σ 1 − σ 3 ) = 56.57 MPa (σ 2 − σ 3 ) = 33.28 MPa (σ 1 − σ 2 ) = 23.28 MPa

σ oct =

1 3

τ oct =

1 3

(σ 1 + σ 2 + σ 3 )= 13.333 MPa

[(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ] 2 = 23.21 MPa 1

______________________________________________________________________________________ SOLUTION (1.83)

( −100 − σ p ) 0 − 80 (a) 0 ( 20 − σ p ) 0 =0 − 80 0 ( 20 − σ p ) Expanding,

( 20 − σ )[(σ + 100)(σ − 20) − 6400] = 0 σ 1 = 60 MPa, σ 2 = 20 MPa, σ 3 = −140 MPa

and

( b ) Apply Eqs. (1.43), (1.44), and (1.45): 1 2

2 2 2 1 τ= ] 86.41 MPa oct 3 [(60 − 20) + (20 + 140) + ( −140 − 60)= 1 −20 MPa σ oct = 3 (60 + 20 − 140) = τ max = 12 (60 + 140) = 100 MPa

______________________________________________________________________________________ SOLUTION (1.84) Octahedral and shearing stresses are given by

2 τ oct = 19 [(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ]

2 τ max = 14 (σ 1 − σ 3 ) 2

Let us say, τ max > τ oct . Then 2

( σ 1 −2σ 3 ) 2 > 19 [(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ]

9 4

(σ 1 − σ 3 ) > [(σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 + (σ 3 − σ 1 ) 2 ]

Subtracting (σ 1 − σ 3 ) from both sides and noting that (σ 1 − σ 3 ) 2

we have

5 4

= (σ 3 − σ 1 ) 2 ,

(σ 1 − σ 3 ) 2 > (σ 1 − σ 3 ) 2 + (σ 2 − σ 3 ) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.84 (CONT.) But Thus,

(σ 1 − σ 3 ) 2 > (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2 5 4

(σ 1 − σ 3 ) 2 > (σ 1 − σ 2 ) 2 + (σ 2 − σ 3 ) 2

(a)

The squares of the difference between σ 1 and σ 3 will always be greater then the sum of the squares of the difference between σ 1 and σ 2 , σ 2 and σ 3 , since σ 1 > σ 2 > σ 3 . Hence, Eq. (a) is true and our assumption is correct. That is

τ max > τ oct

______________________________________________________________________________________ SOLUTION (1.85) From Solution of Problem 1.69:

= σ 1 24.747 = MPa σ 2 8.48 = MPa σ 3 2.773 MPa Applying Eqs. (1.43) and (1.44):

τ oct = 13 [( 24.747 − 8.48) 2 + (8.48 − 2.773) 2 + ( 2.773 − 24.747) 2 ]

1 2

= 9.31 MPa

and

σ= oct

Therefore

1 3

(24.747 + 8.48 + 2.773) = 12 MPa 1

poct = (9.312 + 122 ) 2 = 15.19 MPa

______________________________________________________________________________________ SOLUTION (1.86) Shearing stress, in terms of principal stresses, is given by

τ 2 = σ 12 l 2 + σ 2 m 2 + σ 32 n 2 − (σ 1l 2 + σ 2 m 2 + σ 3n 2 ) 2 2

(a)

We substitute n = 1 − m − l into Eq. (a), calculate its derivatives with respect to l and m , and equate these derivatives to zero: ∂τ ∂l

= l[(σ 1 − σ 3 )l 2 + (σ 2 − σ 3 )m 2 − 12 (σ 3 − σ 1 )] = 0

(b)

∂τ ∂m

= m[(σ 1 − σ 3 )l + (σ 2 − σ 3 )m − 12 (σ 2 − σ 3 )] = 0

(c)

One solution is l = m = 0 . Solutions for the direction cosines of planes for which τ is a maximum of minimum can also be found as follows. Take l = 0 :

Eq. (c) gives m = ±

Take m = 0 :

Eq. (c) gives l = ±

There are, in general, no solutions of Eqs. (b) and (c) in which l and m are both different from zero, for this case the expressions in brackets cannot both vanish. (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.86 (CONT.) By the above procedure we can form the following table. Direction cosines for planes of τ max and τ min

±1

± 12

m = 0 ±1 n = ±1 0

0 0

± 12 ± 12

0 ± 12

± 12 0

The first three columns define the planes for τ min , where τ = 0 . The last three columns give planes through each principal axes bisecting the angles between the two other principal axes. Substituting the latter direction cosines into Eq. (a), we have

(τ 23 ) max = ± σ 2 2−σ 3

(τ 13 ) max = ± σ 1 −2σ 3

(τ 12 ) max = ± σ 1 −2σ 2

Similarly, introducing the direction cosines given in the above table into Eq. (1.37), we obtain the normal stresses associated with the maximum shearing stresses:

σ 12 ' = σ +2σ 1

σ 13 ' = σ +2σ 1

σ 23 ' = σ +2σ

______________________________________________________________________________________ SOLUTION (1.87)

1 (σ x + σ y ) 2 1 = (100 + 20)= 60 MPa 2 σ −σ y 2 2 ( x ) + τ xy R = 2

σ avg =

τ (MPa) σavg =60

100 − 20 2 ( ) + 602 2 = 72.1 MPa

= σ (MPa)

σ1

σ3

σ 1= σ avg + R= 132.1 MPa

σ 3 = −12.1 MPa (a)

σ 2 = 30 MPa

σ 1 = 132.1 MPa

σ 3 = −12.1 MPa

1 (σ 1 − σ 3 ) 2 1 = [132.1 − (−12.1)] = 72.1 MPa 2

(τ max )a =

(b)

σ3 = −30 MPa σ1 = σ2 = −12.1 MPa 132.1 MPa (τ max )a =

1 1 (σ 1 = [132.1 − (= −30)] 81.1 MPa −σ3) 2 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.88) τ (MPa)

σ2

-50

100+ 60 2

o r 45

σ (MPa)

σ1

(100,-20)

= 80

From Mohr’s circle, we have

r = 202 + 202 = 28.3 MPa = σ 1 108 = MPa σ 2 51.7 MPa

σ3 = −50 MPa θp ' = 22.5o ______________________________________________________________________________________ SOLUTION (1.89) τ (MPa)

-60

σ2

80 2

= 40

σ1

σ (MPa)

r (80,-40) Referring to Mohr’s circle:

402 + 402 = 56.57 MPa

θ p ' = 22.5o = σ 1 96.57 = MPa σ 2 16.57 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (1.90) ( a ) In the yz plane: 50 MPa

σ=25

15 MPa

15 σ3

C r

τ(MPa)

O σ1

σ (MPa)

50 We have

r= Thus

252 + 152 = 29.15 MPa

σ 1 =r − σ =29.15 − 25 =4.15 MPa σ 3 =−r − σ =−29.15 − 25 =−54.15 MPa σ 2 = −20 MPa

( b ) Using Eqs. (1.43) and (1.44):

= τ oct

[(4.15 + 20) 2 + (−20 + 54.15) 2 + (−54.15 − 4.15) 2 ] 2 = 23.92 MPa 1 σ oct = −23.33 MPa 3 (4.15 − 20 − 54.15) = 1 3

From Eq. (1.45),

τ max = 12 (4.15 + 54.15) = 29.15 MPa

______________________________________________________________________________________ SOLUTION (1.91) 2

It is noted that l + m + n = 1 . Applying Eq. (1.37), we have

σ= 35( 133 ) − 14( 131 ) − 28( 139 ) = −12.38 MPa

Equation (1.39), substituting the given data and the direction cosines determined above, gives

τ = 26.2 MPa

Surface tractions.

Equations (1.48) give

p= σ= 35( 133= ) 16.81 MPa x 1l py = σ 2m = −14( 131 ) = −3.88 MPa

pz = σ 3n = −28( 139 ) = −23.30 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 1.91 (CONT.) Check: p

= px2 + p y2 + pz2 = σ 2 + τ 2 = 840 (MPa) 2

Observe that Approach I is more conveniently leads to results. ______________________________________________________________________________________ SOLUTION (1.92)

σ oct=

1 3

τ oct=

1 3

We have l=

(40 + 25 + 15)= 26.667 MPa 1

[(40 − 25) 2 + (25 − 15) 2 + (15 − 40) 2 ] 2= 10.274 MPa

m= n= 1

3 . Note that l 2 + m 2 + n 2 = 1 .

Surface tractions. Equations (1.48) give

σ= p= 40( 1= ) 23.09 MPa 1l x 3

1 = p y σ= 25(= ) 14.43 MPa 2m 3 1 = pz σ= 15( = ) 8.66 MPa 3n 3

Check:

p 2 = px2 + p y2 + pz2 = σ 2 + τ 2 = 816.7 (MPa) 2 End of Chapter 1

CHAPTER 2 SOLUTION (2.1) ( a ) Yes. Eqs. (2.12) are satisfied. ( b ) No. Eqs. (2.12) are not satisfied. ______________________________________________________________________________________ SOLUTION (2.2) Apply Eqs. (2.4):

ε x = 2c

(a)

u AB = v AD

γ xy = 2c( x + y )

∫ ε dx= 4=c 0.4 mm = ∫ ε dy = − 6c = −11.25c = −1.125 mm x

Thus,

(b)

ε y = −6cy

2 y2 2 1

LA ' B ' = 2000.4 mm

LA ' D ' = 1498.875 mm

γ xy = 2c(1 + 12 )= 300 µ

( c ) We have and

v A = c(12 − 3 × 14 ) = 0.25c

u A = c( 2 × 1 + 14 ) = 2.25c xA' = 1 + 2.25c = 1000.225 mm

y A' = 0.5 + 0.25c = 500.025 mm

______________________________________________________________________________________ SOLUTION (2.3) Equations (2.4), for the given displacement field, yield [ε ij ] :

0 −y 2   2x  0 2z ( 2 y − x ) 2 c   2z − y 2 ( 2 y − x ) 2  At point (0,2,1), we have [ε ij ] :

0 −100   0  0 200 200  µ   −100 200 200  ______________________________________________________________________________________ 48 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (2.4) First two of Eqs. (2.4) give

ε x = 2a0 xy 2 + a1 y 2 + 2a 2 xy

ε y = b0 x 2 + b1 x

γ xy = co x 2 y + c1 xy + c2 x 2 + c3 y 2 Equation (2.11):

( 4a0 + 2a1 ) + ( 2b0 ) = 2c0 x + c1

2( 2a0 − c0 ) x + 2( a1 + b0 ) − c1 = 0

This is satisfied if x ≠ 0 :

2a0 − c0 = 0, c0 = 2 a 0 c1 = 2( a1 + b0 ) 2( a1 + b0 ) − c1 = 0,

______________________________________________________________________________________ SOLUTION (2.5) Equation (2.11) yields

2a1 + 12 y 2 + 2b1 + 12 x 2 = 3c1 ( x 2 + y 2 ) + c1c2

Solving,

c1 = 4

c2 = 12 ( a1 + b1 )

______________________________________________________________________________________ SOLUTION (2.6) The change in length of bar AB is

−0.2 mm δ AB = ε AB LAB = (−400 ×10−6 )(500) = From triangles B’BF and EE’F:

250 mm B’ 0.2 mm B

F D’

0.2 3 = = , x 46.875 mm x 750 − x

500 mm

δD

From triangles DD’F and EE’F:

203.125

3 mm =

δD

Thus E’

ε= CD

702.125 , δ D 0.8667 mm = 3

δD

= LCD

0.8667 = 1733.4 µ 500

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.7) C

∆LAB 2.5 = − 1800 LAB = −1389 µ

ε AB = 3.0 m

2.4 m

∆LBC

α A

2.5 mm

= ε BC

1..8 m

−∆LBC −0.0025cos α = 3.0 LBC

−0.0025(1.8 3.0) = −500 µ 3.0

______________________________________________________________________________________ SOLUTION (2.8)

2.8 = 1400 µ 2000 −1.3 −1300 µ εy = = 1000 γ xy = 0

(a) = εx

(b)

∠ ACB = 90o 1.0014 = 90.1547 o 0.9987 o o γ= 90 − 90.1547 = −0.1547 o = −2.7 ×10−3 rad = −2700 µ ' B ' 2 tan −1 ∠ A ' C=

______________________________________________________________________________________ SOLUTION (2.9) ( a ) Equations (2.4) give

ε= x ε= y

and

∂u ∂x

∂v ∂y

= 667 µ

0.175 − 0.075 150 0.025 − ( −0.05) 100

= 750 µ

[ −0.125 − ( −0.05)] 0 − 0.075 = −1250 µ γ xy = 100 + 150

( b ) Equation (2.16) is therefore

ε 1, 2 = 667+2 750 ± [( 667−2 750 ) 2 + 6252 ] or When

ε1 = 1335 µ

1 2

ε 2 =µ 82

o −1250 θ p = 12 tan −1 667 −750 = 43.1

and stresses are substituted into Eq. (2.14a), we obtain ε= x'

82 µ . Thus,

θ p ' ' = 43.1

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.10) Use Eq. (2.16) with the strains obtained in Example 2.1:

ε 1, 2 = 1250−22000 ± [( 1250+22000 ) 2 + 750 2 ] or

ε1 = 1415 µ

−2165 µ ε2 =

Apply Eq. (2.15):

= θp

1 2

= tan −1 12501500 12.39o − ( −2000)

1 2

Substituting this angle and the stresses, Eq. (2.14a), gives 1415 µ . Thus,

θ p ' = 12.39 o

______________________________________________________________________________________ SOLUTION (2.11) We have, using Eqs. (2.3):

∂u 0.004 ε= x ∂= 20= 200 µ x ∂v −0.003 εy = −250 µ 12 = ∂y =

and

∂v ∂u γ xy = −1000 − 500 = −1500 µ ∂x + ∂y =

Then, Eq. (2.16) yields

ε 1, 2 = 200−2 250 ± [( 200+2 250 ) 2 + ( −750) 2 ] or

ε1 = 758 µ

1 2

ε2 = −808 µ

Apply Eq. (2.15):

o −1500 θ p = 12 tan −1 200 + 250 = −36.65

For this angle, Eq. (2.14a) gives = ε x'

758 µ . Therefore,

θ p ' = −36.65o ______________________________________________________________________________________ SOLUTION (2.12) y A x”

x’

20 mm θ2

B 12 mm

θ1

We have, from geometry:

AC = QB = 23.32 mm

θ1 = 30.96o

θ 2 = 149.04 o

Equation (2.14a) leads to

300 + 500 ε x' = + 300−2500 cos 61.92o + 100sin 61.92o = 441 µ 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.12 (CONT.) Thus,

∆ QB = 441µ(23.32) = 0.01 mm

Similarly,

ε x" = 400 − 100 cos 298.08o + 100sin 298.08o = 265 µ

and

∆ AC = 265 µ(23.32) = 0.006 mm

______________________________________________________________________________________ SOLUTION (2.13) y x’ 30 mm A B x”

θ2

15 mm θ1

We find, from geometry:

AC = QB = 33.54 mm θ1 = 26.56o θ 2 = 153.44 o

Equation (2.14a) gives Hence,

400 + 200 ε x' = + 400−2 200 cos 53.12o − 150sin 53.12o = 240 µ 2

∆ QB = 240 µ(33.54) = 0.008 mm

In a like manner, and

300 + 100 cos 306.88o − 150sin 306.88o = 480 µ ε x" =

∆ AC = 480 µ(33.54) = 0.016 mm

______________________________________________________________________________________ SOLUTION (2.14) x’ x” C 60 mm B x’’’ θ3 40 mm 30o θ2 x D A We obtain, from geometry:

= AC 96.73 = mm BD 32.29 mm o θ 2 = 11.93 θ 3 = 141.73o From Solution of Problem 1.42:

σ x = 3.464τ 0

σy =0

τ xy = τ 0

Thus, (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.14 (CONT.) 3.464(70×106 )

= εx

= 1212 µ

200(109 )

εy = −0.3(1212) = −364 µ 2(1+ 0.3)(70×106 )

γ= xy

= 910 µ

200(109 )

1212 −364 ε x' = 1212 + 12122+364 cos 60o + 455sin 60o =µ 2 = ∆ AB 1212= µ (40) 0.05 mm

(a)

ε x" = 424 + 788cos 23.86o + 455sin 23.86o = 1328.7 µ 1328.7 µ(96.73) ∆= = 0.13 mm AC

(b)

Similarly,

424 + 788cos 283.46o + 455sin 283.46o = 164.92 µ ε x ''' = = ∆ BD 164.92(32.29) = 0.005 mm

ε 1, 2 = 12122−364 ± [( 12122+364 ) 2 + 4552 ]

ε1 = 1334 µ

1 2

ε2 = −486 µ

Substitution of

θ p ' = 12 tan −1 1212910+364 = 15o

into Eq. (2.14a) yields 1334 µ . ______________________________________________________________________________________ SOLUTION (2.15) ( a ) We have

γ max = 400 − 200 = 200 µ

Maximum shearing strain occurs on a plane oriented at 45o from the plane of principal strains. γ

(106 )

100

(ε x , 12 γ xy ) 60o

200

400

ε (10 6 )

(ε y ,− 12 γ xy ) Figure (a) ( b ) From Mohr’s circle Fig. (a):

εx = 300 + 100 cos 60o = 350 µ

εy = 300 − 100 cos 60o = 250 µ

γ xy = −(400 − 200) sin 60o = −173 µ ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.16) We have u B = 3 mm and vB = 1.5 mm . ( a ) Take: Hence,

Thus,

u = c1 xy

v = c2 xy c1 = 500(10 −6 ) c2 = 250(10 −6 )

3(10 −3 ) = c1 (3 × 2) , 1.5(10 −3 ) = c2 (3 × 2) , u = 500(10 −6 ) xy

( b ) Using Eqs. (2.4),

ε x = 500 µy

v = 250(10 −6 ) xy

ε y = 250 µx

γ xy = 250 µ(2 x + y )

which satisify Eq. (2.11): the strain field is possible. At point B, we thus have

(ε x ) B =µ 1000

( c ) We obtain θ = tan

−1 2 3

(ε y ) B = 750 µ

( γ xy ) B =µ 2000

= 33.69 o .

Equation (2.14a) is therefore

ε x' = 875 + 125cos 67.38o + 1000sin 67.38o = 1846 µ

______________________________________________________________________________________ SOLUTION (2.17) (-300,450)

ε2

2θ p '

(-900,-450)

ε1

(106 )

ε (10 6 )

-600

Figure (a) Refer to Mohr’s circle (Fig. a):

ε1,2 =−600 ± [(−300) 2 + (450) 2 ] or and or

ε1 = −59 µ

1 2

ε2 = −1141 µ

o 450 ( 2 ) 2θ p ' ' = tan −1 −( 900 −300 ) = −56.31

θ p ' = 61.85o

______________________________________________________________________________________ 54 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (2.18) γ

6 2 (10 )

(300,450)

2θ p '

ε2

ε1

ε (10 6 )

(900,-450)

600

Figure (a) Referring to Mohr’s circle shown in Fig. (a), we obtain

ε 1, 2 = 600 ± [(300) 2 + ( 450) 2 ] from which and or

ε1 = 1141 µ

1 2

ε 2 =µ 59

o 450 ( 2 ) 2θ p ' ' = tan −1 900 −300 = 56.31

θ p ' = −61.85o

______________________________________________________________________________________ SOLUTION (2.19) Referring to Fig. P2.19,

Hence,

u A = −0.0005 = c1 (3 × 1 × 2), c1 = −83.3(10 −6 ) v A = 0.0003 = c2 (6), c2 = 50(10 −6 ) w A = −0.0006 = c3 (6), c3 = −100(10 −6 ) u = −83.3(10 −6 ) xyz

v = 50(10 −6 ) xyz

w = −100(10 −6 ) xyz

( a ) Using Eqs. (2.4), we thus have

50 µxz ε= εx = −83.3 µyz εz = −100 µxy y (−83.3 xz + 50 yz ) µ γ xy = γ xz = (−100 yz − 83.3 xy ) µ

γ yz =(50 xy − 100 xz ) µ This forgoing expressions satisfy Eqs. (2.12): the strain field is possible. Substitute x=3 m, y=1 m, and z=2 m into the above equations to obtain strains at A, [ε ij ] :

 −167 −200 −225  −200 300 −225 µ    −225 −225 −300  (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.19 (CONT.) ( b ) Let, for example, x’-axis lie along the line from A to B (Fig. P2.19). The direction of cosines of AB are

m1 = 0

l1 = − 313

n1 = − 213

Thus, Eq. (2.18a):

ε x ' = ε x l12 + ε z n12 + γ xz l1n1 = −167( 139 ) − 300( 134 ) − 450( −133 )( −132 ) = −416 µ

( c ) Let y’-axis be placed along AC (Fig. P2.19). The direction cosines of AC are

m 2 = −1

l2 = 0

Equation (2.18b) is therefore

n2 = 0

γ x ' y ' = γ xy l1m2 + γ yz n1m2 = −400( −133 )( −1) − 450( −132 )( −1)

= −582 µ

Negative sign shows that the angle BAC has increased. ______________________________________________________________________________________ SOLUTION (2.20) We now have (Fig. P2.19):

Hence,

u A = 0.0006 = c1 (3 × 1 × 2), c1 = 100(10 −6 ) v A = −0.0003 = c2 (6), c2 = −50(10 −6 ) w A = −0.0004 = c3 (6), c3 = −66.7(10 −6 ) u = 100(10 −6 ) xyz

v = −50(10 −6 ) xyz

( a ) Applying Eqs. (2.4), we obtain

εy = −50 µxz

= ε x 100 µyz

γ xy = (100 xz − 50 yz ) µ γ yz = (−50 xy − 66.7 xz ) µ

w = −66.7(10 −6 ) xyz

εz = −66.7 µxy

γ xz = (−66.7 yz + 100 xy ) µ

These expressions satisfy Eqs. (2.12): the strain field is possible. Introducing x=3 m, y=1 m, and z=2 m into the above equations we find strains at A, [ε ij ] :

250  200  250 −300   83.3 −275

83.3 −275  µ −200 

( b ) Let, for instance, x’-axis lie along the line from A to B (Fig. P2.19). The direction of cosines of AB are

l1 = − 313

m1 = 0

n1 = − 213

(CONT.) ______________________________________________________________________________________ 56 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 2.20 (CONT.) Therefore, Eq. (2.18a):

ε x ' = ε x l12 + ε z n12 + γ xz l1n1 = 200( 139 ) − 200( 134 ) − 167( −133 )( −132 ) = −154 µ

( c ) Let y’-axis be placed along (Fig. P2.19). The direction cosines of AC are

m 2 = −1

Equation (2.18b) is thus

l2 = 0

n2 = 0

γ x ' y ' = γ xy l1m2 + γ yz n1m2 = 500( −133 )( −1) − 550( −132 )( −1)

= 111 µ

Positive sign means that the angle BAC has decreased. ______________________________________________________________________________________ SOLUTION (2.21) ( a ) Applying Eqs. (2.21),

J1 = 200 − 100 − 400 = −300 µ

and

J 2 =(−2 − 8 + 4 − 9 − 4 − 25)(104 ) =−44(104 ) (µ) 2

200 300 J3 = 300 −100 200 500

( b ) Table of direction cosines:

x' y' z'

3 2 −1 2 0

200 500 = 58(106 ) (µ)3 −400 z

12 0 3 2 0 0 1

Thus, using Eqs. (2.18a),

ε x ' = ε x l12 + ε y m12 + γ xy l1m1 = 200( 23 ) 2 − 100( 12 ) 2 + 600( 23 )( 12 ) = 385 µ

( c ) Use Table B.1 (with σ → ε and τ → γ

598 µ ε1 = (d)

2 ):

ε2 = −126 µ

ε3 = −772 µ

γ max = 598 + 772 = 1370 µ

______________________________________________________________________________________ 57 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (2.22) ( a ) Applying Eqs. (2.21),

J1= 400 + 0 + 600= 1(103 ) µ

and

J 2 = (0 + 24 + 0 − 1 − 4 − 0)(104 ) = 19(104 ) (µ) 2

400 J3 = 100

100 0

0 −200 = −22(106 ) (µ)3

−200

600

( b ) Using Eq. (2.18a),

ε x' = 400( 23 ) 2 + 200( 23 )( 12 ) = 387 µ ( c ) Use Table B.1 (with σ → ε and τ → γ

ε1 = 664 µ

ε2 = 416 µ

2 ):

ε3 = −80 µ

γ max = 664 + 80 = 744 µ

(d)

______________________________________________________________________________________ SOLUTION (2.23) Use Table B.1 (with σ → ε and τ → γ and

2 ):

ε1 = 1807 µ

ε2 = −228 µ

ε3 = −679 µ

l1 = 0.6184

m1 = 0.5333

n1 = 0.5772

______________________________________________________________________________________ SOLUTION (2.24) Referring to App. B (with σ → ε and τ → γ and

2 ), we obtain ε1 = 350 µ ε 2 = 162 µ ε 3 = −488 µ l1 = −0.4356, m1 = −0.884, n1 = 0.1691

______________________________________________________________________________________ SOLUTION (2.25) Nominal strain

= ε0

= 333 µ

0.025 75

Nominal stress

= σ0

9(10 ) = 79.577 MPa

π (0.012)2 4

Modulus of elasticity

= E

79.577(106 )

= 238.97 GPa 333(10−6 )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.25 (CONT.) True strain

ε= ln(1 + 0.000333) = 333 µ

True stress

σ =79.577(1 + 0.000333) =79.603 MPa

______________________________________________________________________________________ SOLUTION (2.26)

ε=

σ=

P A

= 2000 µ

0.10 50

16(103 )

= 141.5 MPa

(π 4)(0.012)2

141.5(106 )

= 70.8 GPa

(a)

(b)

∆δ= νε d= 0.33(2000 ×10−6 )(12)= 7.92(10−3 ) mm e = ε (1 − 2ν ) = 2000(10 −6 )(1 − 2 × 0.33) = 6.80(10 −4 )

σ ε

2000(10−6 )

______________________________________________________________________________________ SOLUTION (2.27) Assume Hooke's Law applies. We have −3

) εy = − 1.5(10 = −300 μ 5

εy

300 εx = 1000 μ −ν = − −0.3 =

Thus,

σ =Eε x =(105 ×109 )(1000 ×10−6 ) =105 MPa σ < σ yp , our assumption is valid.

Since So

2 P σ= A (105)(π 4)(5) 2.062 kN = =

______________________________________________________________________________________ SOLUTION (2.28) We obtain

LAC = LBD = 162 + 162 = 22.63 mm − 22.63 ε x = ∆LL = 22.5922.63 = −1768 μ AC

= εy (a)

∆LBD LBD

= 442 μ

22.64 − 22.63 22.63 6

σx 195(10 ) E= − −1768(10 110 GPa −6 = εx = )

( b ) ν=

εy εx

( c= ) G

110 2(1+ 0.25)

= 0.25

442 −1768

= 44 GPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.29) P =

12(103 ) 250 1.4

12(103 )(5×103 )

( a ) We have Areq =.

σ all

Areq=.

PL δE

and

= 67.2 mm 2 = 75 mm 2

4(200×103 )

Since 75 > 67.2 mm ,

= d (b) = k

AE L

4(75)

= 9.77 mm. π

4A = π

(75×10−6 )(200×109 ) 5

= 3(103 ) kN m

______________________________________________________________________________________ SOLUTION (2.30) The cross-sectional area:

A= Also

(D2 − d 2 ) =

π D2 4

[1 − ( Dd ) 2 ] = 0.589 D 2

= 53.33 mm 2

8 150(103 )

Equating these,

= D 2 90.543 = mm 2 , D 9.52 mm It follows that

8(400) PL δ= = 0.833 ×10−3 m = 0.833 mm AE = (53.33)(72×10 ) 3

= 9.6 kN mm

8000 0.833

______________________________________________________________________________________ SOLUTION (2.31)

(a)

For the solution described, we have ε x = 0 and σ z = 0

It follows that

ε x = E1 [σ x −ν (σ y + σ z )] = E1 [σ x −ν (σ y + 0)]

or (b)

σ x = νσ y = −ν p

(a)

ε y = E1 [σ y −ν (σ x + σ z )] Thus,

ε y = E1 [σ y −ν (σ x + 0)]

Substituting Eq.(a),

ε y = 1−Eν σ y = − 1−Eν p 2

(c)

ε z = E1 [σ z −ν (σ x + σ y )] Insulting Eq.(a):

1 εz = E [0 −ν (νσ y + σ y )]

Thus,

1+ν εz = − ν (1E+ν ) σ y = E p

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.32) d 5 kN

5 kN L

2 mm

5(103 ) 20(103 ) = = 160 ×106 2 2 πd 4 πd = d min 0.0063 = m 6.3 mm

σ= x

Also, or

160 ×106 = 210 ×109 0.002 L Lmin = 2.625 m

E =

______________________________________________________________________________________ SOLUTION (2.33) (a) Axial stress in the bar σ= s

σ= σ is a

= = 126 MPa σ ε= 600(210) s Es Hence (b)

2 P σ= A 126[ (40= )] 158.3 kN = 4

Axial strain in aluminum equals

ε= a Therefore

σa

= Ea

126(106 ) = 1,800 µ 70(109 )

= δ ε a La + ε s Ls

= [1,800(0.5) + 600(1.5)]10−6 = 1.8 mm ______________________________________________________________________________________ SOLUTION (2.34)

A= 2(π × 402 4) = 2.513 mm 2 (a)

P 600(103 ) = = 239 MPa A 2513(10−6 ) Since σ < σ yp , the result is valid.

σ=

Thus,

(b)

239(106 ) = = 1195 µ E 200(109 ) ∆L= Lε = 5(103 )(1195 ×10−6 ) = 5.98 mm

ε=

εt = −νε = −0.3(1195)(10−6 ) = −359 µ ∆d =−359(10−6 )(40) = −0.014 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.35)

ε=

P −150(103 ) = = = −758 µ E AE 70(109 )(π 4)(0.12 − 0.082 )

(a)

∆L =ε L =−758(10−6 )(0.4 ×103 ) = −0.303 mm

(b)

∆ = D ν (ε= D) 0.3(758)(10−6 )100 = 0.023 mm

(c) = ∆t ν (ε= t ) 0.3(758)(10 )20 = 0.0045 mm ______________________________________________________________________________________ SOLUTION (2.36) −6

(a) According to assumption 1, the rubber is in triaxial stress:

σx = σz = − p, Strains are: ε= x

σy = − = − 2 π 2 πd 4

ε= 0. The first of Eqs. (3.17) gives z 1 E

0 =[σ x −ν (σ y + σ z )] εx =

σy

4Q − p) πd2 4ν Q Solving, p = π d 2 (1 −ν )

0 = p −ν ( −

σx σz

(b) Substituting the data,

4(0.3)(5 ×103 ) = p = 1.091 MPa (C ) π (0.05) 2 (1 − 0.3) ______________________________________________________________________________________ SOLUTION (2.37) p=160 MPa= -σ

σ x = σ y = σ z = −σ

d=250 mm

= Vo (a)

4 3 4π 3 πr (125 ) 8.18(106 ) mm3 = = 3 3

εx = − E1 [σ −ν (σ + σ )] = − σE (1 − 2ν )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.37 (CONT.)

−160(106 ) (1 − 0.6) = = −914 µ 70 ×109 ∆d =ε x d =−914(10−6 )250 = −0.229 mm

Decrease in circumference:

π (∆d ) = −0.229π = −0.72 mm

(b)

∆V = (1 − 2ν )ε xVo = (0.4)(−914)(10−6 )(8.18 ×106 ) = −2291 mm3

______________________________________________________________________________________ SOLUTION (2.38) γ

(a)

500 − 100 = 200 µ 2 1 75 tan −1 7o = θp ' = 2 300 θ s " = 45 + 7 = 52o = ε'

(µ)

ε'

ε1

ε2 O

2θ p '

ε (µ)

(500, -75)

ε1 = 200 + 309 = 509 µ ε2 = −309 + 200 = −109 µ

509 µ

y’

R = 752 + 3002 =309 µ = 2= γ max R 618 µ

109 µ

220 µ

x’ 7o (b)

x’

200 µ

618 µ

52o

0.3 0.7 (γ max )t = 509 + 171 = 680 µ

(500 − 100) = ε 3 == εz − −171 µ Thus,

______________________________________________________________________________________ SOLUTION (2.39) γ 2

(µ)

1 (ε x + ε y ) 2 1 = (480 + 800) =640 µ 2

= ε avg

480

ε2

R C

ε1

ε (µ)

2 (640 − 800) 2 + 560 = 582 µ

640

(CONT.) ______________________________________________________________________________________ 63 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 2.39 (CONT.) (a)

ε= ε avg ± R 1,2

ε1 = 640 + 582 = 1222 µ ε 2 =640 − 582 =58 µ (b)

R 2(582) γ max = 2= = 1164 µ

(c)

(γ max )t = ε1 − ε 3 = ε1= − 0 1222 µ

______________________________________________________________________________________ SOLUTION (2.40) (a)

(b)

(c)

P 25(103 ) σ= = = 69.4 MPa x A (0.06 × 0.006) σ x 69.4(106 ) = = 347 µ ε= x 200(109 ) E εy = −νε x = −(0.3)(347 ×10−6 ) = −104 µ 1 1 (ε x + ε y ) + (ε x − ε y ) cos 2θ 2 2 o For θ = −30 : 1 1 ε x ' = (347 − 104) + (347 + 104) cos(−60o ) = 234 µ 2 2 1 1 ε y ' = (347 − 104) − (347 + 104) cos(−60o ) =8.8 µ 2 2

ε x'=

For θ = −30 : o

γ x' y' = −(ε x − ε y ) sin 2θ = −(347 + 104) sin(−60o ) = 391 µ

______________________________________________________________________________________ SOLUTION (2.41) o 0= , θb 45o , and θ c = 90o , Eqs (2.44) give 1 (ε x + ε x + γ xy ) ε a= ε x , ε = ε y , ε= c b 2 ε x = ε a , ε y = ε c , γ xy = 2ε b − (ε a + ε c )

Substituting = θa

Thus, we have

ε x =−800 µ,

ε y =400 µ,

(P2.41)

γ xy =2(−1000) − (−800 + 400) =−1600 µ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.41 (CONT.) Thus,

−800 + 400 1200 2 1600 2 ± (− ) + (− ) = −200 ± 1000 2 2 2 −1200 µ ε1 =µ 800 , ε2 = −1600 tan −1[ ] 26.6o = θ p 12= −800 − 400 ε x ' = −200 − 600 cos 53.2o − 800sin 53.2o = −1200 µ = ε 2 = ε1,2

Thus,

θ p " = 26.6o y’

1200 µ

800 µ

x’ 26.6o x

______________________________________________________________________________________ SOLUTION (2.42)

900 µ (θ a = 0o ), εa = Thus

340 µ (θb = εb = −60o ),

εc = −80 µ (θ c = −120o )

ε a = ε x cos 2 θ a + ε y sin 2 θ a + γ xy sin θ a cos θ a 900(10−6 ) = ε x cos 2 0o + ε y sin 2 0o + γ xy sin 0o cos 0o , ε x = 900 μ

Likewise,

ε b = ε x cos 2 θb + ε y sin 2 θb + γ xy sin θb cos θb 340(10−6= ) ε x cos 2 (−60o ) + ε y sin 2 (−60o ) + γ xy sin(−60o ) cos(−60o ) 340(10−6 ) = 0.25ε x + 0.75ε y − 0.443γ xy

and,

(1)

ε c = ε x cos 2 θ c + ε y sin 2 θ c + γ xy sin θ c cos θ c −80(10−6 )= ε x cos 2 (−120o ) + ε y sin 2 (−120o ) + γ xy sin(−120o ) cos(−120o ) −80(10−6 ) = 0.25ε x + 0.75ε y + 0.443γ xy

(2)

Subtract Eq. (2) from Eq. (1):

420 μ = −0.886 γ xy

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.42 (CONT.) So

γ xy = −474 μ

and then from Eq. (1):

ε y = −126.6 μ

γ 2

εx ε avg

γ xy

εy

1 (ε x + ε y ) 2 1 = (900 − 126.6) = 386.7 μ ε 2 474 2 12 = R [(900 − 386.7) 2 + (− ) ] 2 = 565.4 μ = ε avg

ε= ε avg ± R 1,2

ε1 = 386.7 + 565.4 = 952.1 μ ε 2 = −178.7 μ tan 2θ p ' =

−474 , 900 + 126.6

θ p ' = −12.4o

______________________________________________________________________________________ SOLUTION (2.43) y

τ xy Tc π 3 J c= = γ xy J G 2 σ x= σ y= 0 ε x= ε y= 0 θ= φ= 30o

= τ xy A

Mohr’s circle for strain:

1 ε avg = (ε x + ε y ) = 0 2

1 2

1 R = γ xy 2

εφ = ε avg + R sin 2θ = γ xy sin 2θ = This gives

τ xy

2φ

Tc = sin 2φ sin 2φ 2G 2GJ

A’ x’

2GJ ε φ c sin 2φ

Substitute the given data:

π (81×109 )(0.045)3 (620 ×10−6 ) (0.045) sin 60o

= 369 kN ⋅ m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.44)

DB = 0.05 2 m

10 MPa D

y’

5 MPa

50 mm

x’ 45o

20 MPa x

( a ) Hooke’s law gives

ε y = (10−0E.3×20 ) = E4

ε x = ( 20−0E.3×10 ) = 17E

γ xy = 5( 2E.6 ) = 13E Then, using Eq. (2.14a) with θ=

ε y' = Thus, = ∆ BD

1 2E

θ +π 2, (17 + 4) − 0 − 213E sin 90o = E4

(ε y ' )( BD) =

0.283 E

( b ) Applying Eqs. (1.18a,c):

o o 30 10 20 MPa σ x' = 2 + 2 cos 90 + 5sin 90 =

15 − 5cos 90o − 5sin 90o = 10 µ σ y' = As before, Hooke’s law yields

ε y ' = (10−0E.3×20 ) = E4

and

= ∆ BD (ε y ' )( = BD)

0.283 E

______________________________________________________________________________________ SOLUTION (2.45) The generalized Hooke’s law gives

ε x = 0 = E1 [σ x − ν (σ y + σ z )]

or Solving, Thus

ε z = 0 = E1 [σ z − ν (σ x + σ y )] σ x − νσ z = νσ y − νσ x + σ z = νσ y σ x = σ z = 1−1ν σ y

σ x = σ z = − 1ν−pν

______________________________________________________________________________________ SOLUTION (2.46) Using Eqs. (2.46): 1 2

ε 1, 2 = 12 {( −100 + 100) ± [( −200) 2 + 100 2 ] } = 12 (0 ± 224) (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.46 (CONT.)

ε1 = 112 µ

ε2 = −112 µ

σ 1, 2 = 200(210 ) [0 ± ( 224 1.3 )] or

σ 1 = 17.2 MPa

and

σ 2 = −17.2 MPa

o θ p = 12 tan −1 −100 200 = −13.28

Equations (2.44),

ε x = εa

ε y = εc

γ xy = 2ε b − ε a − ε c

Equation (2.14a) gives Thus,

ε x' = 0 − 100 cos(−26.56o ) + 50sin(−26.56o ) = −112 µ

θ p ' ' = −13.28o

______________________________________________________________________________________ SOLUTION (2.47) We have

ε x + ε y = ε a + ε c = 1200 µ

and the first two of Eqs. (2.44):

1000 = ε x cos 2 ( −15o ) + ε y sin 2 ( −15o ) + γ xy sin( −15o ) cos( −15o ) − 250 = ε x cos 2 30o + ε y sin 2 30o + γ xy sin 30o cos 30o

These may be written

ε y = 1200 − ε x

1000 = 0.933ε x + 0.067ε y − 0.25γ xy − 250 = 0.75ε x + 0.25ε y + 0.433γ xy Solving,

ε x =522 µ

ε y =678 µ

γ xy =−1873 µ

______________________________________________________________________________________ SOLUTION (2.48) ( a ) We have

εc = εy = −50 µ

First two of Eqs. (2.44):

400 = ε x ( 43 ) − 50( 14 ) + γ xy ( 12 )( 23 ) 300 = ε x ( 43 ) − 50( 14 ) + γ xy ( − 12 )( 23 ) Solving,

ε x =µ 483

γ xy =µ 115

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.48 (CONT.) Applying Eq. (2.16),

ε 1, 2 = 4832−50 ± [( 4832+50 ) 2 + 582 ] or

ε1 = 489 µ

Thus,

1 2

ε2 = −56 µ

γ max = ε1 − ε 2 = 545 µ

( b ) Using Eq. (3.11b) of Chap.3, Hence,

ε z = − 1ν−ν (483 − 50) = −217 µ = ε 3 (γ max )t = ε1 − ε 3 = 706 µ

______________________________________________________________________________________ SOLUTION (2.49) From Eqs. (2.35), (2.37), and (2.38), we have 9

(10 ) ν = 2200 − 1 = 0.25 ( 80×10 ) 9

and

λ = 0.251×.25200×0(.105 ) = 80(109 ) e = 200 + 300 = 500 µ

Then, Eqs. (2.36) lead to the following stress components, [τ ij ] :

72 16 0  16 88 64  MPa    0 64 40  ______________________________________________________________________________________ SOLUTION (2.50) Equation (2.35) yields

= ν

200(109 )

= − 1 0.25

2(80×109 )

Then, introducing the given data into the generalized Hooke’s law, Eqs. (2.34), we calculate the following strain components, [ε ij ] :

31.25  81.25 −25  −25 −43.75 62.5  µ   31.25 62.5 50 

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.51) Using Eq. (2.35),

G =

70(109 ) 2(1+ 0.3)

= 26.92 GPa

For x=1 m, y=2 m, and z=4 m, we obtain [τ ij ] :

34 48 8  48 5 1 MPa    8 1 5 Then, Eqs. (2.34) yield [ε ij ] :

 443 892 149  892 −96 19  µ   149 19 −96  ______________________________________________________________________________________ SOLUTION (2.52) Substituting x=3/4 m, y=1/4 m, and z=1/2 m into Eqs. (d) of Example 1.2, we have [τ ij ] :

 −0.359 2.625 0.234   0.234 0.875 0  MPa    0.234 0 0.125  Equation (2.35) gives,

= G

200(109 ) 2(1+ 0.25)

= 80 GPa

Applying Hooke’s law, we compute the strain components [ε ij ] :

 −3 33 3  33 5 0  µ    3 0 0  ______________________________________________________________________________________ SOLUTION (2.53) We have σ z = 0 . Using Eq. (2.34), Hooke’s law:

ε x=

1 E

(σ x −νσ y )=

1 70×103

(60 − 903 )= 1286 µ

ε y=

1 E

(σ y −νσ x )=

1 70×103

(90 − 603 )= 1571 µ

εz = − νE (σ x + σ y ) = − 2101×10 (60 + 90) = −714 µ 3

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.53(CONT.) (a)

∆ AB = ε y b = (1571×10−6 )(400) = 0.628 mm

(b)

e = ε x + ε y + ε z = 1286 + 1571 − 714 = 2143 µ V0 = 300 × 400 ×10 = 1.2 ×106 mm3 ∆V = eV0= (2143 ×10−6 )(1.2 ×106 )= 2,571.6 mm3

______________________________________________________________________________________ SOLUTION (2.54) ( a ) Using generalized Hooke’s law, 10 [−60 − 0.3(−50 − 40)] = εx = −165 µ 200(10 ) 6

εy =

1 200(103 )

[−50 − 0.3(−60 − 40)] = −100 µ

1 εz = [−40 − 0.3(−60 − 50)] = −35 µ 200(10 ) 3

Thus,

∆a =aε x =−0.04125 mm ∆b =bε y =−0.02 mm ∆c =cε z =−0.00525 mm e= εx +εy +εz = −300 µ

(b) and

∆V = e(abc) = −2250 mm3

______________________________________________________________________________________ SOLUTION (2.55) Applying generalized Hooke’s law,

ε= x

106 70(109 )

[70 − 13 (−30 − 15)] = 1214 µ

1 εy = [−30 − 13 (70 − 15)] = −690 µ 70(10 ) 3

1 εz = [−15 − 13 (70 − 30)] = −405 µ 70(10 ) 3

Thus,

∆a= aε x= 0.1821 mm ∆b =bε y =−0.0690 mm ∆c =cε z =−0.0304 mm e = ε x + ε y + ε z = 119 µ

(b) and

∆ = V e(abc = ) 133.88 mm3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.56) We have 200(109 ) 2(1+ 0.3)

= 76.92 GPa

G =

0.3×200(109 ) 1.3(0.4)

= λ

= 115.38(109 )

ε 1 = 5c

Let

ε 2 = 4c

We then obtain e=12c.

ε 3 = 3c

Using the first of Eqs. (2.36),

σ 1 = 2Gε 1 + λe

140(106 ) = 2(76.92 ×109 )(5c) + 115.38(109 )(12c)

This yields

c = 65(10−6 )

Hence,

ε1 = 325 µ

ε2 = 260 µ

ε3 = 195 µ

Now applying Eqs. (2.36), we calculate the principal stresses as follows:

σ 1 = 50 + 90 = 140 MPa σ 2 = 40 + 90 = 130 MPa σ 3 = 30 + 90 = 120 MPa

Therefore or

σ 3 : σ 2 : σ 1 = 120 :130 :140

σ 3 : σ 2 : σ 1 = 1:1.083 :1.167

______________________________________________________________________________________ SOLUTION (2.57)

ε x = σE = ∂∂ux

(a)

ε y = − νσE = ∂∂yv

Integrating; Then

u = σE x + f ( y ) ∂u ∂y

gives

v = − νσE y + g ( x )

(a)

+ ∂∂vx = 0

∂f ( y ) ∂y

∂g ( x ) ∂x

+ ∂g∂(xx ) = 0 or

Integration leads to

g ( x ) = cx + d

Equations (a) are thus

u = σE x − cy + e

= − ∂f∂(yy ) = c

f ( y ) = −cy + e

v = − νσE y + cx + d

Boundary conditions u(0,0)=0 and v(0,0)=0 result in c=d=e=0. Thus

u = σE x

v = − νσE y

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ (CONT.)

ε x = σE = const.

(b) Hence

ε y = − νσE = const.

ε x = u x and ε y = −ν y . Therefore u = σE x and v = − νσE y

______________________________________________________________________________________ SOLUTION (2.58) Equations (2.34) become 2

ε x = cy +Eνcx

ε y = − cx E+νcy

Integrating,

Given:

u( x, y ) = ∫ ε x dx = 3cE (3 y 2 x + νx 3 ) + g1 ( y )

(1)

v ( x, y ) = ∫ ε y dy = 3cE (3 y 2 x + νy 2 ) + g 2 ( x )

(2)

τ xy = 0

(3)

τ xy = ∂∂uy + ∂∂vx = 0

From Eq. (1), ∂u ∂y

= dgdy1 + 2 Ec xy

(a)

Equation (2) leads to ∂v ∂x

= dgdx2 − 2 Ec xy

dg1 dy

= 2 Ec xy + a1

(b)

Substituting Eqs. (a) and (b) into Eq. (3): dg 2 dx

= −2 Ec xy + a1

from which, after integration,

g1 ( y ) = Ec xy 2 + a1 y + a 2

(4)

g 2 ( x ) = − Ec yx 2 − a1 x − a 2

Constants a1 and a 2 are obtained upon satisfying the prescribed boundary conditions. Substitution of Eqs. (4) into Eqs. (1) and (2) yields the solution for displacement field. ______________________________________________________________________________________ SOLUTION (2.59) Equations (2.39) and (2.35) give

K = 3(1−E2ν ) = 23G(1(−12+νν ))

Also,

K = 3(1−E2ν ) = 3(12+Eν ()(1+1ν−2)ν ) = 3E(1(+3νν +)(11−−22νν)) = (1+ν ν)(E1−2ν ) + 3(1E+ν ) = λ + ( 2G3 )

Equations (2.39) and (2.35) yield 2

G = 3 K2 ((11+−ν2ν) ) = ( 3−1E+ 2(ν1−) 2(ν1−) 2ν ) = E {[E3 (1(1−−22νν)])−1} = 9E

3E 2 3 E (1− 2ν )

1 3 1− 2ν

−E

= 93KKE− E

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.59 (CONT.) Equations (2.35) and (2.39) give,

E = 2G (1 + ν )

and G ( 3+ 2 G ) λ +G

E 2 (1+ν )

E = 3K (1 − 2 E )

[3λ + 1+Eν ]

λ + 2 (1E+ν )

The above expression, after substituting λ from Eq. (2.38) and simplifying, reduce to E. From formula (2.39), or

(1 − 2ν ) = 3EK ν = 12 − 6EK = 3K6 K− E

We also have or

(1 + ν ) = 2EG

G = 2EG − 1

ν = 2EG − 1

This expression is written as

ν = 3 K (21G−2ν ) − 1 = 32KG − 62KGν − 1

from which or

ν + 62KGν = ν (1 + 26νG ) = 32KG − 1

ν = 23( K3 K−+2GG ) Finally, we write

λ = 2 (1+ν2ν)(E1−2ν ) = 12−ν2Gν

(1 − 2ν )λ − 2νG = 0

This yields,

ν = 2 ( λλ+G )

______________________________________________________________________________________ SOLUTION (2.60) Equations (2.4) yield,

ε y = − νγ ( Ea − x )

ε x = γ ( aE− x )

γ xy = − νγEy + νγEy = 0

The stresses are therefore,

σ x = 1−Eν (ε x + νε y ) = γ ( a − x ) 2

σ y = 1−Eν (ε y + νε x ) = 0 2

τ xy = Gγ xy = 0 At x=0 and x=a, we have

σ x = γa

σx =0

(CONT.) ______________________________________________________________________________________ 74 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 2.60 (CONT.) Applying Eqs. (1.48) at x=a, we obtain

 p x  0 0 0 1  0         p y  = 0 0 0 0 = 0  p  0 0 0 0 0  z     

This is, boundary conditions at x=a are satisfied. At y = ± b , stresses are and

σ x = γ (a − x )

σ y = τ xy = 0

 p x  ( x − a )γ     py  =  0 p   0  z 

0 0  0  0     0 0 ± 1 = 0  0 0  0  0

We see that boundary conditions at y = ± b are also satisfied. ______________________________________________________________________________________ SOLUTION (2.61) Equations (2.4) give,

ε x = − νγEz , γz

εz = E ,

ε y = − νγEz

γ xy = γ yz = γ xz = 0

(a)

Equations (2.12) are satisfied by the above strains. Equations (2.36) and (a) yield,

σ x = 2Gε x + λ (ε x + ε y + ε z ) = 0

and Note that

σy =0

σ z = γz

Fx = Fy = 0

and

γ xy = τ yx = τ xz = 0 Fz = −γ

Thus, the first of Eqs. (1.14) are satisfied, and the third leads to

0 + 0 + ∂∂σzz + Fz = 0 or γ − γ = 0

At ends, since x’ and z are parallel, n=cos(x’,z)= ± 1. Thus, boundary conditions (1.48) at z=L:

px = 0 + 0 + 0 = 0

py = 0 + 0 + 0 = 0

p z = 0 + 0 + γL(1) = γL as required. Similarly, at z=0:

px = 0

py = 0

p z = γ (0)( −1) = 0

Therefore, all equations of elasticity are satisfied. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.62)

Using Eq. (2.59), with P1 = P1 + P2 : 2

P2 ) L U = ( P12+ AE

P1 P2 From Example 2.8, with P1 = P2 = P :

U v = 12σ E ( AL) = 3PAEL 2

Hence,

5σ 5P L U d = 12 E ( AL ) = 3 AE 2

U = U v + U d = 2 PAEL 2

______________________________________________________________________________________ SOLUTION (2.63) We have 2

L 4) + P2 E( 3( 2LA4)) = 85 U 1 U 2 = P 2(AE

U 1 = 2PAEL And

L 8) + P2 E( 7( 3LA8)) = 125 U 1 U 3 = P 2(AE

Comparison of these results show that strain energy decreases as the volume of the bar is increased, although all three bars have the same maximum stress. ______________________________________________________________________________________ SOLUTION (2.64) Stress field is described by

σ x = σ y = σ z = − p,

( a ) Equation (2.37) reduces to

e = − 3(1−E2ν ) p ;

τ xy = τ xz = τ yz = 0

1−( 2 3)] p − 0.005 = − 3[110 (109 )

p = 550 MPa

( b ) Equation (2.52) becomes 2

U 0 = 21E ( p 2 + p 2 + p 2 ) − νE ( p 2 + P 2 + p 2 ) = 3 p 2(1E−2ν )

3(550×106 )2 2(110×109 )

= (1 − 23 ) 1.375 MPa

Thus,

U = U 0V0 = 1.375(10 6 )( 43π × 0.153 ) = 19.439 kN ⋅ m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.65) Substituting the given data into Eq. (2.52), we have 3

(10 ) (3000 + 2400 + 2000) U 0 = 2 (10200 ) (60 2 + 50 2 + 40 2 ) − 0.3200 3

= 8.15 kPa Thus,

U = U 0 ( abc )

= 8.15(10 3 )(0.25 × 0.2 × 0.15) = 61.125 N ⋅ m ______________________________________________________________________________________ SOLUTION (2.66) Equation (2.59) leads to 2

U n = P 2( 3AEL 4 ) + 2P( n(2LA)4E) = 8PAEL (3 + n12 ) 2

We have, Thus, Hence,

= 1+43nn2 2PAEL

for n = 1 :

U 1 = 2PAEL

for n = 12 :

U 1 2 = 7U4 1

for n = 2 :

U 2 = 1316U1

U1 2 > U1

and

U 2 < U1

______________________________________________________________________________________ SOLUTION (2.67) Axial strain energy

UC = Square Part:

= US Requirement::

P L = 2 AE

Use Eq. (2.59), circular Part:

P2 L 2P2 L = πd2 π d 2E 2( )E 4

P 2 L P 2 (3L 4) 3P 2 L = = 2 AE 2a 2 E 8a 2 E

2 P 2 L 3P 2 L U C U= ; = S π d 2 E 8a 2 E d 16 = a 3π ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.68) (a)

U = 21E ∫ dx ∫ σ 2 dA = 21E ∫ dx ∫ ( PA + MyI ) 2 dA Since

∫ ydA = 0 , this becomes: U =∫ +∫ =U +U P 2 dx 2 AE

Here

M 2 dx 2 EI

a b 2 2 U b = 2 1EI  ∫ M AD dx + ∫ M DB dx   0  0 a b = 2 1EI  ∫ ( − ML0 x ) 2 dx + ∫ ( ML0 x ' ) 2 dx '  0  0 2

= 6MEIL0 2 ( a 3 + b 3 ) 2

U = 2PAEL + 6MEIL0 2 ( a 3 + b 3 ) 2

Thus,

( b ) Substituting the given data: 3 2

3 2

) ( 2×10 ) ( 0.027 + 0.729 ) U = 2 ( 7.(58××1010−3))((701.2×10 9 + ) 6 ( 70×109 )( 0.075×0.13 12 )(1.2 ) 2

= 0.0731 + 0.8 = 0.8731 N ⋅ m ______________________________________________________________________________________ SOLUTION (2.69)

U = ∑ 2TJGL = T π( L 42 ) + T π( L 42 ) 2

d1 ) G

d2 )G

= 8πTGL ( d14 + d14 ) 2

(a)

TL = Tπ( L 2 ) + Tπ( L 2 ) φ = ∑ JG 32

d14G

d 24G

= 16πGTL ( d14 + d14 ) 1

(b)

Solving T from Eq. (b) and substituting into Eq. (a) result in Eq. (P2.69). ______________________________________________________________________________________ SOLUTION (2.70) We have for segment AB, T AB = 3T and for segment BC , TBC = T (a)

U AB = ( 2TJGL ) AB = π (91T.4 d(1)4.2Ga )16 = 44.977 πTd 4aG 2

and

U BC = ( 2TJGL ) BC = 16πdT4Ga 2

The total energy is thus

U = 60.981 πTd 4aG 2

( b ) Substituting the given data into the above equation, 3 2

×10 ) (0.5) = = U 60.981 π(1.4 2.831 kN ⋅ m (0.02)4 (42×109 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.71) p

( Lx − x 2 ) b h

pL 2

p 2

= M

pL 2

The maximum bending moment occurs at the midspan: 2

pL σ max = M I c ( pLbh8)(12h 2 ) = 34 bh max

Maximum strain energy density, 2

2 4

9p L U 0,max = σ2max E = 32 Eb 2 h 4

Using Eq. (2.63), we obtain L

U = 2 1EI ∫ ( 2p ) 2 ( Lx − x 2 ) 2 dx 0

p 2 L5 240 EI

2 5

L = 20pEbh 3

It is required to find c: or U 0,max = cU V

c = U 0,max UV

2 4

p L bhl = 458 c = 329 Eb 2 4 h p 2 L5 20 Ebh 3

Thus, and

U 0,max = 458VU

______________________________________________________________________________________ SOLUTION (2.72) P

A a We have

M AB = Px

x L a

x’

M BC = Pax L

Pa L

U AB = 2 1EI ∫ ( Px )dx = P6 EIa

2 3

0 a

2 P a L U BC = 2 1EI ∫ ( Pax L ) dx ' = 6 EI 2 2

Therefore

U = U AB + U BC = P6 EIa ( L + a ) 2 2

______________________________________________________________________________________ SOLUTION (2.73)

A P 2

− MLo

x L/2

M AC = ( P2 − MLo ) x

B x’ L/2

P 2

+ MLo

M BC = ( P2 + MLo ) x '

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.73 (CONT.) We have L

2 U AB = 2 1EI ∫ M AB dx = ( P2 − ML0 ) 2 ( 48LEI ) + M 0 ( P2 − ML0 )( 8LEI ) + M4 EI0 L 2

Thus

U BC = 2 1EI ∫ M BC dx ' = ( P2 + ML0 ) 2 ( 48LEI ) 2

U = U AB + U BC 2

PM 0 L M0 L P L = 96 EI + 16 EI + 6 EI 2 3

______________________________________________________________________________________ SOLUTION (2.74) Equation (2.66) yields

τ oct = 13 [( −19 − 4.6) 2 + ( 4.6 + 8.3) 2 + ( −8.3 + 19) 2 1

+6(4.7 2 + 6.452 + 11.82 )] 2 = 15.11 MPa Applying Eqs. (2.65) and (2.64), respectively,

= U 0d

= (15.11) 2 2226.71 Pa

3 4(76.9×109 )

and

U 0v =

( −19 + 4.6 −8.3)2 (1012 )

171.76 Pa = 18(166.67×109 )

It follows that U0 d U0 v

= 12.96 ≈ 13

______________________________________________________________________________________ SOLUTION (2.75) Dilatational stress is

= σm

(200 −50 + 40) 3

= 63.33 MPa

Distortional stresses are then

σ x − σ m = 200 − 63.33 = 136.7 MPa σ y −σm = −50 − 63.33 = −113.3 MPa σ z − σ m =− 40 63.33 = −23.33 MPa

Dilatational and deviator stress matrices are, respectively,

0  63.33 0  0 63.33 0  MPa   0 0 63.33 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 2.75 (CONT.) and

20 10  136.7  20 −113.3 0  MPa   10 0 −23.33

Next, substitute the above deviator stresses into Eq. (1.33). Then, solve the resulting equation for principal deviator stresses, using Table B.1:

σ 1d = 138.8 MPa

σ 2 d = −23.9 MPa

σ 3d = −114.9 MPa

______________________________________________________________________________________ SOLUTION (2.76) Only existing stresses are: σ x = σ

τ xy = τ

Here,

= τ

2(20×103 )

4(15×103 )

= σ We have

= 58.95 MPa

π r3

π (0.06)3

= 88.42 MPa

π r3

π (0.06)3

G = 25E

K = 23E

Equations (2.64) and (2.65): 2

) (10 ) U 0 v = 18σ K = 12σ E = (8812.42( 200 ) 2

= 3.258 kPa

U 0 d=

5 12 E

(σ 2 + 3τ 2 )=

5(103 ) 12(200)

(7818.1 + 10, 425.3)

= 38.007 kPa Thus,

U 0 = U 0 v + U od = 41.265 kPa

______________________________________________________________________________________ SOLUTION (2.77) Principal stresses are 1 2

σ 1, 2 = σ2 ± [ σ4 + τ 2 ] , 2

σ3 = 0

and

τ oct = 13 ( 2σ 2 + 6τ 2 )

1 2

Equations (2.64) and (2.65) are therefore

U 0 v = 181K (σ 12 + σ 22 ) = 18σ K = 1−6 2Eν σ 2 2

2 U 0 d = 43G τ oct = 13+Eν (σ 2 + 3τ 2 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (2.78) Only existing stress components are: σ x = σ where,

σ = πPr

τ xy = τ

τ = π2rT

The area properties are:

A = πr 2

J = π2r

Thus, Eqs. (2.64) and (2.65) become

U 0 v = 18σ K = 1−6 2Eν σ 2 = 12σ E 2

2 5σ 5τ = 12 U 0 d = 43G τ oct E + 4E 2

The components of strain energy are

U v = ∫ U 0 v dV = 121E ∫ PA dx = 12PπrL2 E 2

U d = ∫ U 0 d dV = 125E ∫ PE dx + 45E ∫ TJ dx 2

= 125πPr 2LE + 25πTr 4LE Total strain energy is therefore

U = 2πLr 2 E ( P 2 + 5 Tr 2 ) 2

End of Chapter 2

CHAPTER 3 SOLUTION (3.1) ( a ) We obtain ∂ 4Φ ∂x 4

= −12 pxy

∂ 4Φ ∂y 4

= 6 pxy

∂ 4Φ ∂x 2∂y 2

Thus, ∇ Φ = −12 pxy + 2(6 pxy ) = 0 and the given stress field represents a possible solution. (b)

= pxy 3 − 2 px 3 y

∂ 2Φ ∂x 2

Integrating twice 3 3

Φ = px6 y − px10 y + f1 ( y ) x + f 2 ( y ) 4

The above is substituted into ∇ Φ = 0 to obtain d 4 f1 ( y ) dy

x + d dyf 2 4( y ) = 0

This is possible only if d 4 f1 ( y ) dy 4

d 4 f2 ( y )

dy 4

We find then

f 1 = c 4 y 3 + c5 y 2 + c 6 y + c 7 f 2 = c8 y 3 + c9 y 2 + c10 y + c11

Therefore,

3 3

Φ = px6 y − px10 y + ( c4 y 3 + c5 y 2 + c6 y + c7 ) x + c8 y 3 + c9 y 2 + c10 y + c11

( c ) Edge y=0:

−a a

−a

px pa t Vx = ∫ τ xytdx = ∫ ( 2 + c3 )tdx =5 + 2c3at

Py = ∫ σ y tdx = ∫ (0)tdx = 0

Edge y=b:

Vx = ∫ (− 32 px 2b 2 + c1b 2 + px2 + c3 )tdx −a

= − pa 3 (b 2 − a5 )t + 2a ( c1b 2 + c3 )t 2

Py =∫ ( pxb3 − 2 px 3b)tdx =0 −a

______________________________________________________________________________________ SOLUTION (3.2) Edge x = ± a :

τ xy = 0 :

− 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0

τ xy = 0 :

− 23 pa 2 y 2 + c1 y 2 + 12 pa 4 + c3 = 0

Adding, ( −3 pa

+ 2c1 ) y 2 + pa 4 + 2c3 = 0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.2 (CONT.)

c1 = 23 pa 2 Edge x = a : σx = 0:

c3 = − 12 pa 4

pa 3 y − 2c1ay + c2 y = 0

c2 = 2 pa 3

______________________________________________________________________________________ SOLUTION (3.3) ( a ) Equations (3.6) become ∂σ x ∂x

∂τ

∂τ xy ∂x

+ ∂yxy = 0

Substituting the given stresses, we have Thus

(b)

c 2 y − 2 c3 y = 0 c 2 = 2 c3

c1 = arbitrary

σ x = c1 y + c2 xy

τ xy = c2 (b 2 − y 2 ) 2

Assume c1 > 0 and c2 > 0 . y

τ xy =

c2 2

(b − y )

σ x = c1 y

τ xy = c2 (b 2 − y 2 ) 2

σ x = ( c1 + c2 a ) y

a ______________________________________________________________________________________ SOLUTION (3.4) Boundary conditions, Eq. (3.6): ∂σ x ∂x

∂τ

+ ∂yxy = 0

∂τ xy ∂x

∂σ

+ ∂yy = 0

or ( 2ab − 2ab) x = 0 ( −2ab + 2ab) y = 0 are fulfilled. However, equation of compatibility:

( ∂∂x 2 + ∂∂y 2 )(σ x + σ y ) = 0 or 4ab ≠ 0 is not satisfied. 2

Thus, the stress field given does not meet requirements for solution. ______________________________________________________________________________________ SOLUTION (3.5) It is readily shown that

∇ 4Φ1 = 0 ∇4Φ 2 = 0

is satisfied is satisfied

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.5 (CONT.) We have 2

σ x = ∂∂yΦ = 2c, 2

σ y = ∂∂xΦ = 2a,

Thus, stresses are uniform over the body. Similarly, for Φ 2 :

σ x = 2cx + 6dy

σ y = 6ax + 2by

τ xy = − ∂∂xΦ∂y = −b 1

τ xy = −2bx − 2cy

Thus, stresses vary linearly with respect to x and y over the body. ______________________________________________________________________________________ SOLUTION (3.6) Note: Since σ z = 0 and ε y = 0 , we have plane stress in xy plane and plane strain in xz plane, respectively. Equations of compatibility and equilibrium are satisfied by

σ x = −σ 0

σ y = −c

σz = 0

τ xy = τ yz = τ xz = 0 We have

εy = 0

(a)

(b)

Stress-strain relations become

εx =

(σ x −νσ y ) E

εz =

−ν (σ x +σ y ) E

εy =

, ,

(σ y −νσ x ) E

(c)

γ xy = γ yz = γ xz = 0

Substituting Eqs. (a,b) into Eqs. (c), and solving

σ y = −νσ 0 2

ε x = − (1σ−νE ) 0

ε z = ν (1+Eν )σ

εy = 0

Then, Eqs. (2.3) yield, after integrating: 2

u = − (1−ν E)σ 0 x

v=0

w = ν (1+νE )σ 0 z

______________________________________________________________________________________ SOLUTION (3.7) Equations of equilibrium, ∂τ

2axy + 2axy = 0

∂σ

ay 2 − ay 2 = 0

∂σ x ∂x

+ ∂yxy = 0,

∂τ xy ∂y

+ ∂xy = 0,

are satisfied. Equation (3.12) gives

( ∂∂x 2 + ∂∂y 2 )(σ x + σ y ) = −4ay ≠ 0 2

Compatibility is violated; solution is not valid. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.8) We have ∂ 2ε x

∂ 2ε y

∂y 2

∂x 2

∂ 2γ xy ∂x∂y

= −2ay

= 2ay

Equation of compatibility, Eq. (3.8) is satisfied. Stresses are

σ x = 1−Eν (ε x + νε y ) = 1−aEν ( x 3 + νx 2 y ) 2

σy =

E 1−ν 2

(ε y + νε x ) =

aE 1−ν 2

( x 2 y + νx 3 )

2 τ xy = Gγ xy = 2 (aE 1+ν ) xy

Equations (3.6) become aE 1−ν 2

(3x 2 + 2νxy ) + 1aE +ν xy = 0

aE 1−ν 2

y 2 + 1−aEν 2 x 2 = 0

These cannot be true for all values of x and y. Thus, solution is not valid. ______________________________________________________________________________________ SOLUTION (3.9)

ε x = ∂∂ux = −2νcx

ε y = ∂∂yv = 2ax

γ xy = ∂∂uy + ∂∂vx = −2cy + 2cy = 0 Thus

σ x = 1−Eν (ε x + νε y ) = 0

τ xy = Gγ xy

σ y = 1−Eν (ε y + νε x ) = 2 Ecx

O 2b

2Eac

x 2Eac

Note that this is a state of pure bending. ______________________________________________________________________________________ SOLUTION (3.10) (a)

σ x = ∂∂yΦ = 0,

σ y = 6 pxy

τ xy = −3 px 2

Note that ∇ Φ = 0 is satisfied. (b)

σ y = 6 pbx y

σx = 0 τ xy = 0 b

τ xy = 3px 2 σx = 0 a

σy =0 ( c ) Edge x = 0 : Edge x = a :

τ xy = 3pa 2 x

τ xy = 3px 2

V y = Px = 0 Px = 0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.10 (CONT.) b

V y = ∫ τ xy tdy = 3 pa 2 bt ↓ Edge y = 0 :

Py = 0

V x = ∫ τ xy tdx = pa 3t → 0

V x = pa 3t ←

Edge y = b : a

Py = ∫ σ y tdx = 3 pa 2 bt ↑ 0

______________________________________________________________________________________ SOLUTION (3.11) 4

( a ) We have ∇ Φ ≠ 0 is not satisfied. 2

σ y = ∂∂xΦ = pya , 2

τ xy = 2p (1 + 4 ax )

σx = 0

σ x = p(1 + ay )

2 2

τ xy = 2pya ( 4a + y ) 2

σ y = 0, τ xy = 0 0 := Vy

Edge x = a :

σy = p py τ xy = − 2a

τ xy = − p ( 42xya+ y )

(b)

σ x = p ( xa + xy ) ,

a py 2

dy ∫= 2 0 2a

1 6

pat= Px 0

V y = ∫ τ xy tdy = 76 pat ↓ 0 a

Px = ∫ σ x tdy = 23 pat → Edge y = 0 :

Vx = 0

Edge y = a :

Py = 0

V x = ∫ τ xy tdx = 23 pat ← 0 a

Py = ∫ σ y ptdx = pat ↑ 0

______________________________________________________________________________________ SOLUTION (3.12) 4

( a ) We have ∇ Φ = 0 is satisfied. The stresses are

σ x = ∂∂yΦ = − bpx (6b − 12 y )

σ y = ∂∂xΦ = 0

τ xy = − ∂∂x∂Φy = 6bpy (b − y ) 2

(CONT.)

______________________________________________________________________________________

______________________________________________________________________________________ 3.12 (CONT.) (b)

σ x = − bpa (6b − 12 y )

τ xy

a ______________________________________________________________________________________ SOLUTION (3.13) The strain-displacement relations:

ε= x

= 2c1 x, ε = y

∂u ∂x

= 2c1 x

∂υ ∂y

(a)

∂u ∂υ (−2c1 y − c2 ) + 2c1 y = γ xy = −c2 ∂y + ∂x =

Compatibility condition (2.11) ∂ 2ε x ∂y

∂ 2ε

y += ∂x 2

∂ 2γ xy ∂x∂y

; = 0+0 0

Inserting Eqs.(a) into Eqs.(3.10b), we obtain

σ x = σ y = 1−Eν 2 Ec1 x, τ xy = −GEc2

(b)

Observe that Eqs.(1.13) are not satisfied. Thus, the displacement field is not a possible solution. ______________________________________________________________________________________ SOLUTION (3.14) Substituting Eqs.(a) into Eq.(3.10a) with Eq. (2.35): 2 3P   εx  0   4 h3 x y  1 ν   1   P [ y3 − 3 y − 1] 1 0 ν −  ε y  =    2 3 h3 2 h γ  E  0 0 2(1 +ν )   3 P 2 2     4 h2 x(h − y )   xy   

Insulting these strains into Eq.(3.8): ∂ 2ε x ∂y

∂ 2ε

∂ 2γ

+ ∂x2y ≠ ∂x∂xyy

We are let to conclude thus that, given stresses are not suited for the elasticity analysis. ______________________________________________________________________________________ SOLUTION (3.15) We have ∂Φ ∂y

= − πP [tan −1 xy + x 2xy+ y 2 ],

∂ 2Φ ∂y 2

= − πP [ x 2 +x y 2 + ( x (+xy2 +) yx2−)22 y x ]

∂Φ ∂x

−y = − Py π x2 + y2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.15 (CONT.) The stresses are thus,

σ x = ∂∂yΦ = − 2πP ( x +x y ) 2

2 2

σ y = ∂∂xΦ = − 2πP ( x xy+ y ) 2

2 2 2

τ xy = − ∂∂x∂Φy = − 2πP ( x x+ yy ) 2

2 2

σx L

2 P πL

τ xy

______________________________________________________________________________________ SOLUTION (3.16) Various derivatives of Φ are: ∂Φ ∂x

= τ40 ( y − yh − hy2 ),

∂ Φ ∂x 2∂y 2

= 0,

∂ Φ ∂x∂y

∂ 2Φ ∂x 2

=0 2

= τ40 (1 − 2hy − 3hy2 )

∂ 2Φ ∂y 2

= 4τ 0h ( −2 x − 6hxy + 2 L + 6hLy2 )

∂ 4Φ ∂x 4

= 0,

∂ 4Φ ∂y 4

(a)

It is clear that Eqs. (a) satisfy Eq. (3.17). On the basis of Eq. (a) and (3.16), we obtain

σ x = 4τ h ( −2 x − 6hxy + 2 L + 6hLy ), 0

σy =0

(b)

τ xy = − τ4 (1 − 2hy − 3hy ) 0

From Eqs. (b), we determine Edge y = h : σy =0

τ xy = τ 0

Edge y = − h :

σy =0

τ xy = 0

x = L:

σ x = 0,

τ xy = − τ4 (1 − 2hy − 3hy )

Edge

It is observed from the above that boundary conditions are satisfied at y = ± h , but not at x = L . ______________________________________________________________________________________ SOLUTION (3.17) 4

For ∇ Φ = 0, e = −5d and a, b, c are arbitrary.

(a) Thus

Φ = ax 2 + bx 2 y + cy 3 + d ( y 5 − 5 x 2 y 3 )

(1) (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.17 (CONT.) ( b ) The stresses:

σ x = ∂∂yΦ = 6cy + 10d ( 2 y 3 − 3x 2 y )

(2)

σ y = ∂∂xΦ = 2a + 2by − 10dy 3

(3)

τ xy = − ∂∂x∂Φy = −2bx − 30dxy

(4)

Boundary conditions:

σ y = −p

τ xy = 0

(at y=h)

(5)

Equations (3), (4), and (5) give

b = −15dh 2

2a − 40dh 3 = − p

∫ σ dy = 0 −h

∫ yσ dy = 0

−h

Equations (2), (4), and (7) yield

(6) h

∫ τ dy = 0 −h

(at x=0)

c = −2dh 2

Similarly

σy =0

give

(8)

τ xy = 0

(at y=-h)

a = 20dh 3

(9)

Solution of Eqs. (6), (8), and (9) results in

b = − 163ph

a = − 4p

(7)

c = − 40ph

d = 80ph3

e = − 16ph3

(10)

The stresses are therefore

σ x = − 320pyh + 8hp ( 2 y 3 − 3x 2 y ) 3

σ y = − 2p − 38pyh − 8pyh

3 3

τ xy = 38pxh (1 − hy ) 2

______________________________________________________________________________________ SOLUTION (3.18) We obtain 2

σ x = ∂∂yΦ = p ( x a−2 y ) 2

σ y = ∂∂xΦ = pya 2

(a)

τ xy = − ∂∂x∂Φy = − 2

2 pxy a2

Taking higher derivatives of Φ , it is seen that Eq. (3.17) is not satisfied. Stress field along the edges of the plate, as determined from Eqs. (a), is sketched bellow.

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.18 (CONT.) y

σy = p

τ xy = 2 p ax

τ xy = 2 p ay

τ xy = 0 σ x = 2 p ay

σ x = p(1 − 2 ay )

a 2

σ y = 0, τ xy = 0

______________________________________________________________________________________ SOLUTION (3.19) The first of Eqs. (3.6) with Fx = 0 ∂τ xy ∂y

= pxy I

Integrating, 2

τ xy = pxy 2 I + f1 ( x )

(a)

The boundary condition, 2

(τ xy ) y =h = 0 = pxh 2 I + f1 ( x ) 2

gives f 1 ( x ) = − pxh

τ xy = −

px 2I

2 I . Equation (a) becomes

(h − y 2 )

(b)

Clearly, (τ xy ) y = − h = 0 is satisfied by Eq. (b). Then, the second of Eqs. (3.6) with Fy = 0 results in ∂σ y ∂y

= p ( h2 −I y )

Integrating, 2

σ y = 2pI y ( h 2 − y3 ) + f 2 ( x )

(c)

Boundary condition, with t = 3I 2h ,

(σ y ) y = − h = − pt = − 2phI ( h 2 − h3 ) + f 2 ( x ) 3

gives f 2 ( x ) = − ph

σy =

p 6I

3I . Equation (c) is thus 2

(3h y − y 3 − 2h 3 )

(d)

This satisfies the condition that (σ y ) y =h = 0 . Note that Eq. (3.12) is not satisfied: the solution obtained does not provide a compatible displacement field. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.20) Substituting the stresses from Eqs. (3.10) into Eqs. (3.6) and taking Fx = Fy = 0 :

∂ε

( ∂∂εxx + ν ∂xy ) + G ∂yxy = 0

E 1−ν 2

( ∂yy + ν ∂∂εyx ) + G ∂xxy = 0

∂ε

∂γ

E 2 (1+ν )

[ 1−2ν ∂∂xν2 + 12−νν ∂∂x∂νy + ∂∂yu2 + ∂∂x∂νy ] = 0

E 2 (1+ν )

[ 1−2ν ∂∂yν2 + 12−νν ∂∂x∂uy − ∂∂x∂uy + ∂∂xν2 ] = 0

The foregoing become

∂γ

E 1−ν 2

∂ 2u ∂x 2

+ 11+−νν ∂∂xu2 − 11+−νν ∂∂x∂vy + ∂∂yu2 = 0

∂ 2v ∂y 2

+ 11+−νν ∂∂yv2 + 11+−νν ∂∂x∂vy + ∂∂yu2 = 0

∂ 2u ∂x 2

+ ∂∂yu2 + 11+−νν ∂∂x ( ∂∂ux + ∂∂yv ) = 0

∂ 2v ∂y 2

+ ∂∂xv2 + 11+−νν ∂∂y ( ∂∂yv + ∂∂ux ) = 0

______________________________________________________________________________________ SOLUTION (3.21) It is readily found that

And

∂τ xy ∂y

∂σ x ∂x

= − pxy I

∂σ y ∂y

= − pI ( −h 2 + y 2 )

= pxy I ∂τ xy ∂x

= − 2pI ( h 2 − y 2 )

Thus, Eqs. (3.6) are satisfied: stress field is possible. We= have xQ 1.5 = m, yQ 0.05 m , and 3

I = 2 th3 = 2 ( 0.043)( 0.1) = 2.67(10−5 ) m 4 3

Substituting the given data, Eqs. (P3.19) yield at Q:

= σx

−10×103 10(2.67×10−5 )

−10×10 (5 ×1.52 + 2 × 0.12 )0.05 + 3(2.67 (1.25 ×10−4 ) ×10−5 ) 3

= −21.12 MPa −10×10 ×1.5 −2.107 MPa τ xy = (0.12 − 0.052 ) = 2(2.67×10 ) 3

−5

−10×10 σ y =6(2.67 (2 × 0.13 − 3 × 0.0005 + 0.053 ) =−0.039 MPa ×10 ) 3

−5

Applying Eq. (2.29), we have

G =

200(109 ) 2(1+ 0.3)

= 76.9 GPa

Hooke’s law is therefore

1 εx = (−21.12 + 0.3 × 0.039)106 = −106 µ 200(10 ) 9

εy=

1 200(109 )

(−0.039 + 0.3 × 21.12)106 = 31.5 µ 6

−2.107(10 ) γ xy = = −27.4 µ 76.9(10 ) 9

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ (CONT.) Principal strains are 2 −27.4 2 ε1,2 =−1062+31.5 ± [( −137.5 2 ) +( 2 ) ]

ε1 = 32.9 µ

We have θ p =

1 2

ε2 = −107 µ

−27.4 tan −1 −106 5.63o = −31.5

1 2

For this angle, Eq. (2.14a) yield ε x ' = −107.4 µ . Thus,

θ p ' = 5.63o ______________________________________________________________________________________ SOLUTION (3.22) Assume ε x = ε y = 0,

σ z = 0,

σ x = σ y = constant

which satisfy Eqs. (3.6) and (3.27). Hooke’s law becomes

ε x = E1 (σ x − νσ y ) + αT1 = 0

(a)

ε y = (σ y − νσ x ) + αT1 = 0 ε z = νE ( −σ x − σ y ) + αT1 = 0 From Eqs. (a) and (b), we obtain σ x = σ y . Therefore, 1 E

(b) (c)

ε x = σE (1 − ν ) + αT1 = 0 x

This yields

σ x = σ y = Eνα−T1

Then, Eq. (c) becomes

ε z = 21να−νT + αT1 We also have: τ xy = τ yz = τ xz = 0 and γ xy = γ yz = γ xz . 1

______________________________________________________________________________________ SOLUTION (3.23) The nonzero strain components are

ε x = ε y = ε z = αT

Compatibility equations reduce to ∂ 2ε

∂ 2γ

∂ 2ε x

+ ∂x 2y = ∂y∂xyx

∂ 2ε y ∂z 2

+ ∂∂yε2z = ∂y∂yzz

∂ 2ε z ∂x 2

+ ∂∂zε2x = ∂∂xγ∂xzz

∂y 2

Adding the above equations and substituting the given data:

2α ( ∂∂xT2 + ∂∂yT2 + ∂∂zT2 ) = 0 2

∂ 2T ∂x 2

+ ∂∂yT2 + ∂∂zT2 = 0 2

(CONT.) ______________________________________________________________________________________ 93 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 3.23 (CONT.) This equation, for a time independent temperature field, has the solution

T = c1 ' x + c2 ' y + c3 ' z + c4 '

which may be written as

αT = c1 x + c2 y + c3 z + c4

______________________________________________________________________________________ SOLUTION (3.24) Stress is σ x = −σ 0 regardless of T1 and still we have ε y = 0 . The second of Eqs. (3.26a) is thus or

ε y = E1 (σ y + νσ 0 ) + αT1 = 0 σ y = −νσ 0 − EαT1

(a)

The first of Eqs. (3.26a) and (a) result in

ε x = − 1−Eν σ 02 + (1 + ν )αT1 2

(b)

Now, Hooke’s law

ε z = − νE (σ x − σ y ) + αT1

leads to

ε z = ν (1E+ν ) σ 0 + (1 + ν )αT1

(c)

Then Equations (2.4) yield, after integration,

u = εxx

v=0

w = εzz

Here, ε x and ε z are given by Eqs. (b) and (c).

______________________________________________________________________________________ SOLUTION (3.25)

Equation (3.27) reduces to d2 dy 2

(σ x + αET ) = 0

from which

σ x = −αET + c1 y + c2 = −αET ( a1 y + a 2 ) + c1 y + c2 Referring to Part (b) of Example 3.2: c1 = c2 = 0 and σ x = −αE ( a1 y + a 2 )

We have

Px = ∫ σ x tdy = −2 Eαhta 2 −h

M z = ∫ σ x tydy = − Eαt a13y + a22y −h

h −h

= − Eαth a1 2 3

The Px and M z are opposite to that shown above.

______________________________________________________________________________________ 94 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (3.26) Assume a stress distribution:

τ xy = τ yz = τ xz = σ y = σ z = 0

σ x = constant

(a)

which satisfy Eqs. (3.6) and (3.27). Then, Hooke’s law becomes

ε x = σE + αT , ε y = ε z = − νσE + αT γ xy = γ yz = γ xz = 0 x

(b,c)

(d)

Due to constraint imposed by the walls and because of the uniformity of the temperature distribution: we take ε x = 0 . Equations (a) and (b) give

σ x = −αET

Equation (c) is then

ε y = ε z = αT (1 + ν ) = constant

The compressive force P on the tube is

Px = σ x A = − EαAt

Substituting the data given

Px = −120(10 9 )(16.8 × 10 −6 ) × (800 × 10 −6 )(100) = −161.3 kN

______________________________________________________________________________________ SOLUTION (3.27) Derivatives of the given stress function are ∂ 2Φ ∂r 2

= 0,

1 ∂Φ r ∂r

Equation (3.40) becomes

= − Pπθr sin θ ,

1 ∂ 2Φ r 2 ∂θ 2

= Pπθr sin θ − 2πPr cos θ

∇ 4 Φ = ( ∂∂r 2 + 1r ∂∂r + r12 ∂∂θ 2 )( − 2πPr cos θ ) 2

After performing the derivatives, we obtain ∇ Φ = 0 . Substituting the derivatives obtained above, into Eqs. (3.32):

σ r = − Pπθr sin θ + Pπθr sin θ − 2πPr cos θ = − 2πPr cos θ Similarly, we find σ θ = 0 and τ rθ = 0 .

______________________________________________________________________________________ SOLUTION (3.28) Refer to Fig. P3.28a. Let ABC = 1 and hence AAB = cos θ , AAC = sin θ .

∑F = 0: x

σ x = σ r cos θ cos θ + σ θ sin θ sin θ − 2τ rθ sin θ cos θ ∑ Fy = 0 : τ xy = σ r cos θ sin θ − σ θ sin θ cos θ + τ rθ cos θ cos θ − τ rθ sin θ sin θ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ (CONT.) Similarly, from Fig. P3.28b:

∑F = 0: y

σ y = σ r sin 2 θ + σ θ cos 2 θ + 2τ rθ sin θ cos θ

Check:

∑F = 0: x

τ xy = σ r sin θ cos θ − σ θ sin θ cos θ + τ rθ (cos 2 θ − sin 2 θ ) Thus, quoted equations are derived. ______________________________________________________________________________________ SOLUTION (3.29) Apply the chain rule (Sec. 3.9): and

∂Φ ∂x

= ∂∂Φr cos θ − 1r ∂∂Φθ sin θ

∂ 2Φ ∂x 2

θ = ∂∂rΦ2 cos 2 θ − 2 ∂∂θ∂Φr sinθ rcosθ + ∂∂Φr sinr θ − 2 ∂∂Φr sinθrcos + ∂∂θΦ2 sinr 2 θ 2 2

(a)

∂ 2Φ ∂y 2

θ = ∂∂rΦ2 sin 2 θ + 2 ∂∂θ∂Φr sinθ rcosθ + ∂∂Φr cosr θ + 2 ∂∂Φθ sinθrcos + ∂∂θΦ2 cosr 2 θ 2

(b)

Similarly,

Adding Eqs. (a) and (b), we have ∂ 2Φ ∂x 2

+ ∂∂yΦ2 = ∂∂rΦ2 + 1r ∂∂Φr + r12 ∂∂θΦ2

By referring to the identity

(c)

+ 2 ∂x∂2∂Φy 2 + ∂∂yΦ4 = ( ∂∂x 2 + ∂∂y 2 )( ∂∂xΦ2 + ∂∂yΦ2 )

∂ 4Φ ∂x 4

and Eq. (c), we can readily write the equation quoted, Eq. (3.40). ______________________________________________________________________________________ SOLUTION (3.30) Equation (3.28) is written as

( drd 2 + 1r drd )( ddrΦ2 + 1r ddrΦ ) + αET ( drd 2 + 1r drd ) = 0 2

or or

d 2Φ dr 2

+ 1r ddrΦ + αET = 0

1 d r dr

( r ddrΦ ) + αET = 0

______________________________________________________________________________________ SOLUTION (3.31) ( a ) Let C = 2 (sin 2α −−2Mα cos 2α ) , and

Φ = C (sin 2θ − 2θ cos 2α ) Various derivatives of Φ are: ∂Φ ∂r

∂Φ ∂θ

= 2C cos 2θ − 2C cos 2α

∂ Φ ∂θ 2

∂ 2Φ ∂r 2

= −4C sin 2θ

(CONT.) ______________________________________________________________________________________ 96 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 3.31 (CONT.) and

2θ ∇ 2 Φ = r12 ∂∂θΦ2 = − 4C sin r2 2

We thus obtain

2θ 2θ ∇ 4 Φ = C[ 8 sinr 42θ − 24 sin + 16 sin ]=0 r4 r4

(b)

2θ σ r = r1 ∂∂θΦ = − 4C sin σθ = 0 r τ rθ = − ∂∂r ( 1r ∂∂Φθ ) = 2rC (cos 2θ − cos 2α ) 2

(c)

Letting α = π 2 : C = − M 2π . It follows that 2θ σ r = 2 Mπsin r

σθ = 0

θ τ rθ = 2rC (cos 2θ + 1) = − 2 Mπcos r 2

where cos θ = (1 + cos 2θ ) 2 ______________________________________________________________________________________ SOLUTION (3.32) 2

Using Eq. (3.48) and Fig. P3.32: π

Fx = ∫ (σ r rdθ ) sin θ = ∫ ( 2πP cos θ sin θ )dθ 2

= 2πP 12 sin 2 θ Similarly,

= πP π

Fy = ∫ π (σ r rdθ ) cos θ = ∫ π ( 2πP cos 2 θ )dθ 2

− 2

= 2πP θ2 + 14 sin 2θ −2π = P 2

______________________________________________________________________________________ SOLUTION (3.33) NOTE: (In P3.33, 3.34, and 3.35), the P, L, and α are constants. It can readily be verified that, the maximum values of the functions in parentheses occur: d dθ

(sin θ cos 3 θ ) = 0 , or tan 2 θ = 13

when θ = ±30

d dθ

(sin 2 θ cos 2 θ ) = 0 , or tan 2 θ = 1

when θ = ±45

o o

Maximum stresses, using Eqs. (3.37) and (3.43):

(σ x ) elast . = L (α + 1Psin 2α )

(a)

(τ xy ) elast . =

P sin θ cos3 θ L (α + 12 sin 2α )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.33 (CONT.) Elementary solution of maximum stresses are P (σ x ) elem. = 2 L tan α ,

(τ xy ) elem. = 0

(b)

( a ) For α = 15 , (α + 12 sin 2α = 0.512) : o

Thus,

(σ x ) elast . = P 0.512 L

at θ = 0

(τ xy ) elast . = P 2.195L

at θ = 15

(σ x ) elem. = P 0.536 L

at any θ

o o

(σ x ) elast . = 1.047(σ x ) elem.

( b ) For α = 60 , (α + 12 sin 2α = 1.48) : o

Thus,

(σ x ) elast . = P 1.48L

at θ = 0

(τ xy ) elast . = P 4.557 L

at θ = 30

(σ x ) elem. = P 3.464 L

at any θ

o o

(σ x ) elast . = 2.341(σ x ) elem.

______________________________________________________________________________________ SOLUTION (3.34) See: NOTE, solution of Prob. 3.33. Substitute α = 30 into Eqs. (a) and (b) of Solution of Prob. 3.33. o

Thus,

(σ x ) elast . = P 0.957 L

at θ = 0

(τ xy ) elast . = P 2.946 L

at θ = 30

(σ x ) elem. = P 1.155L

at any θ

(σ x ) elast . = 1.207(σ x ) elem.

______________________________________________________________________________________ SOLUTION (3.35) See: NOTE, solution of Prob. 3.33.

We have r = L cos θ , hmn

2 = c = L ⋅ tan α , and I = 2c 3 3 .

Equations (3.43) give

cos θ (σ x ) elast . = LF(αsin−θ1 sin 2α ) 2

θ cos θ (τ xy ) elast . = LF(sin α − 1 sin 2α ) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.35 (CONT.) Elementary solution:

(σ x ) elem. = 3FL 2c 2 ,

(τ xy ) elem. = 3F 4c

( a ) For α = 15 , (α + 12 sin 2α = 0.512) : o

Thus,

(σ x ) elast . = 19.43F L

at θ = 15

(τ xy ) elast . = 5.21F L

at θ = 15

(σ x ) elem. = 20.89 F L

at θ = 15

(τ xy ) elem. = 2.8F L

at θ = 0

o o o

(σ x ) elast . = 0.93(σ x ) elem. (τ xy ) elast . = 1.86(τ xy ) elem.

( b ) For α = 60 , (α − 12 sin 2α = 0.614) : o

Thus,

(σ x ) elast . = 0.529 F L

at θ = 30

(τ xy ) elast . = 0.407 F L

at θ = 45

(σ x ) elem. = 0.5F L

at θ = 60

(τ xy ) elem. = 0.433F L

at θ = 0

o o

(σ x ) elast . = 1.058(σ x ) elem. (τ xy ) elast . = 0.94(τ xy ) elem.

______________________________________________________________________________________ SOLUTION (3.36) With x = r ⋅ cos θ , Eqs. (3.50) become

σ x = −( 2πPr ) cos 3 θ

σ y = −( 2πPr ) sin 2 θ cos θ τ xy = −( 2πPr ) sin θ cos 2 θ Substituting,

prdθ 2p 3 2 dσ x = − π2r ( cos θ ) cos θ = − π cos θdθ

dσ y = − 2πp sin 2 θ , Integrating,

dτ xy = − πp sin 2θdθ

θ2

σ x = − 2πp ∫ cos 2 θdθ = − 2pπ [2(θ 2 − θ1 ) + (sin 2θ 2 − sin 2θ1 )] θ1

σy =−

p 2π

[2(θ 2 − θ1 ) − (sin 2θ 2 − sin 2θ1 )]

τ xy = 2pπ [cos 2θ 2 − cos 2θ1 ] ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.37) APPROACH (a):

( drd 2 + 1r drd )( ddr 2f1 + 1r dfdr1 ) = 0 2

We have

(d)

d dr

( ddr 2f1 ) = ddrf31 ,

d dr

( 1r dfdr1 ) = − r −2 dfdr1 + r −1 ddr 2f1

d2 dr 2

( 1r dfdr1 ) = r23 dfdr1 − 2r ddr 2f1 + 1r ddr f31

d2 dr 2

( ddr 2f1 ) = ddr 4f1 2

1 d r dr

( ddr 2f1 ) = 1r ddrf31

1 d r dr

( 1r dfdr1 ) = 1r ( − r12 dfdr1 + 1r ddr 2f1 )

Then, Eq. (d) becomes d 4 f1 dr

+ 2r ddrf31 − r12 ddr 2f1 + r13 dfdr1 = 0

(d’)

The first equation of Problem 3.37 may be written as: 1 d r dr

{r drd [ r −1 drd ( r dfdr1 )]} = 0

1 d r dr

{r drd [ r −1 dfdr1 + ddr 2f1 ]} = 0

1 d r dr

{r[ − r −2 dfdr1 + r −1 ddr 2f1 + ddrf31 ]} = 0

1 d r dr

{− r −1 dfdr1 + ddr 2f1 + r ddr f31 } = 0

1 r

{r −2 dfdr1 − r −1 ddr 2f1 + 2 ddrf31 + r ddr 4f1 } = 0

1 df1 r 3 dr

− r12 ddr 2f1 + 2r ddrf31 + ddr 4f1 } = 0

which is the same as Eq. (d’). Now let us integrate the expression: 1 d r dr

{r drd [ 1r drd ( r dfdr1 )]} = 0

r drd [ 1r drd ( r dfdr1 )] = c1 1 d r dr

( r dfdr1 ) = c1 ln r + c2

r dfdr1 = c1 ∫ r ln rdr + c2 ∫ rdr r dfdr1 = c1 [ r2 ln r − r4 ] + c2 r2 + c3 2

df1 dr

= c1r ln r + c2 r 2 + c3 ln r + c4

f1 = c1r 2 ln r + c2 r 2 + c3 ln r + c4

Expression (e) of Section 3.12 may be treated in a like manner. APPROACH (b): Letting t = ln r , we have df1 dr

= dfdt1 drdt = 1r dfdr1

d 2 f1 dr 2

= r12 ( ddt 2f1 − dfdt1 )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.37 (CONT.) d 3 f1 dr 3

= r13 ( ddt 3f1 − 3 ddt 2f1 + 2 dfdt1 )

d 4 f1

= 1r ( ddt 4f1 − 6 ddt 3f1 + 11 ddt 2f1 − 6 dfdt1 )

Substituting these derivatives into Eq. (d’), we obtain: d 4 f1 dt

− 4 ddt 3f1 + 4 ddt 2f1 = 0

This is an ordinary differential equation with constant coefficients. It has a solution

f1 = c1r 2 ln r + c2 r 2 + c3 ln r + c4

In a like manner, it can be shown that, 2

( drd 2 + 1r drd − r42 )( ddrf22 + 1r drdf22 − 4rf22 ) = 0 2

is solved to yield Eq. (g) of Section 3.12. ______________________________________________________________________________________ SOLUTION (3.38) (a)

σ0

2σ 0

σ r1 = σ2 [(1 − ar ) + (1 + 3ra − 4ra ) cos 2θ ] 0

σ θ 1 = σ2 [(1 + ar ) − (1 + 3ra ) cos 2θ ] 0

τ rθ 1 = − σ2 (1 − 3ra + 2ra ) sin 2θ 0

and

σ r 2 = σ2 [(1 − ar ) + (1 + 3ra − 4ra ) cos 2(θ + 90o )] 0

σ θ 2 = σ2 [(1 + ar ) − (1 + 3ra ) cos 2(θ + 90o )] 0

τ rθ 2 = − σ2 (1 − 3ra + 2ra ) cos 2(θ + 90o ) 0

We have, by superposition:

σ r = σ r1 + σ r 2 σ θ = σ θ1 + σ θ 2 Hence, at r=a and θ = π 2 , σ r1 = 0 σ r2 = 0 σ θ 1 = 3σ 0 σ θ 2 = −σ 0 τ rθ 1 = 0 τ rθ 2 = 0

τ rθ = τ rθ 1 + τ rθ 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.38 (CONT.) lead to the solution:

σr = 0

σ θ = 2σ 0

τ rθ = 0

( b ) Referring to the results of part ( a ), we write

σ r1 = 0 σ r2 = 0 σ θ 1 = 3σ 0 σθ2 = σ 0 τ rθ 1 = 0 τ rθ 2 = 0 Thus, σ r = 0 σ θ = 4σ 0 τ rθ = 0

______________________________________________________________________________________ SOLUTION (3.39) We have

d D

= 13

Then, from Fig. D.9: K ≈ 2.3 . Hence 3

180(10 ) P 2.3 (150 σ= K= = 207 MPa max −50)20 A

______________________________________________________________________________________ SOLUTION (3.40)

σ max 130 = = 1.625 σ nom 80 From Fig. D.1: r d = 0.25 . Then

= K

gives or

D = 2r + d

= 40 2(0.25d= ) + d 1.5d d = 26.7 mm r = 6.67 mm

______________________________________________________________________________________ SOLUTION (3.41) For

r = 0.15 : d D = 2r + d ; 40 = 2(0.15d ) + d = 1.3d

or d = 30.76 We thus have

D = 1.3 d Figure D.1 gives K ≈ 1.7. Hence, 6 = σ max 250(10 = ) 1.7

Pall (20)(30.76)

Pall = 90.5 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.42)

1 kN ⋅ m TBC =

2 kN ⋅ m TAB =

τ max = 2T π c3 yields

Then,

2(1×103 ) = 79.6 MPa τ BC = π (0.02)3 = τ AB Thus

2(2 ×103 ) = 81.5 MPa π (0.025)3 = τ max K= τ AB (1.6)(81.5) = 130.4 MPa

______________________________________________________________________________________ SOLUTION (3.43) (a)

We have

D 40 r 2 = = 1.14 = = 0.057 d 35 d 35 Hence, by Fig. D.7, K ≈ 1.6 .

We have

J = So

(d )4 (35) 4 147.3(103 ) mm 4 = = 32 32

Tc 100(0.0175) ] 19 MPa = τ yp K= 1.6[ = J 147.3(10−9 )

______________________________________________________________________________________ SOLUTION (3.44) We have

D 28 r 4 = = 1.4 = = 0.2 d 20 d 20 From Fig. D.7 , K ≈ 1.28 Tc 16T 6 K ; 250(10= ) 1.28[ ] τ max = τ= all J π (0.016)3 or

= T 157.1 N ⋅ m

______________________________________________________________________________________ SOLUTION (3.45) 5 r = 1.5, = 0.2 d= 25 Figure D.1, K = 1.72 . 210 1.4 max Hence σ nom = σ1.72 = = 87.2 MPa 1.72 and Pall = Aσ nom = (25 ×10)(87.2) = 21.8 kN

D d

37.5 25

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.46) At a section through B

= M B 400(0.3) = 120 N ⋅ m M Bc 120(0.02) σ= = = 37.5 MPa nom 1 I (0.012)(0.04)3 12 (a)

r 5 = = 0.125 d 40

D 60 = = 1.5 : d 40

K ≈ 1.65

(Fig. D.2)

K ≈ 1.41

(Fig. D.2)

= σ max 1.65(37.5) = 61.9 MPa r 10 D 60 = = 0.25 = = 1.5 : d 40 d 40 = σ max 1.41(37.5) = 52.9 MPa

(b)

______________________________________________________________________________________ SOLUTION (3.47) 4.8 r = 36 24 = 1.5, d = 24 = 0.2 Figure D.1, K = 1.72 . D d

Hence

and

σ max

2 = 210 = 61.05 MPa 1.72 Pall == Aσ nom (30 ×15)(61.05) = 27.5 kN

σ nom =

1.72

______________________________________________________________________________________ SOLUTION (3.48) At the notch and hole:

P 15 ×103 = = 20.8 MPa σ nom = ( D − d h − 2r )t (0.108 − 0.018 − 2 × 0.009)(0.01) For the notch : (see Fig. D.1):

D 108 = = 1.2 d 90

r 9 = = 0.1, K= 1.78 t d 90

= σ max K= 1.78(20.8) = 37 MPa tσ nom For the hole (see Fig. D.9):

d h 18 = = 0.167, = K 2.5 D 108 Hence = σ max K= σ nom 2.5(20.8) = 52 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.49)

σθ θ

σa Without hole:

σ θ = pd 2t

σ a = pd 4t

With hole:

2σ a

σa

σa (1)

(2)

We use Eq. (3.55b), with r=a.

σ θ 1 = σ2 [(1 + aa ) − (1 + 3aa ) cos 2θ ] a

= σ a [1 − 2 cos 2θ ]

σ θ 2 = 2σ a [1 − 2 cos 2(θ + 90o )] For θ = 0 : o

σ θ 1 = −σ a For θ = π 2 : σ θ 1 = 3σ a

σ θ 2 = 6σ a

σ θ 2 = −2σ a o Therefore, superposing the results at θ = 0 : σ θ = 5σ a = 5 pd 4t at θ = π 2 : σ θ = σ a = pd 4t

______________________________________________________________________________________ SOLUTION (3.50) ( a ) We have D/d=1.1 and r/d=0.05. Then, we find from Figs. D.7, D.8, and D.6 that

= K t 1.64 = K b 2.2 = K a 2.3

Then, Eqs. (j) of Example 3.4 yield 3

) ×10 ) σ x =2.3 π50(10 + 2.2 4(10 =4.42 MPa (0.2) π (0.2) 2

2(20×10 ) = τ xy 1.64 = 2.61 MPa π (0.2)3

Equation (i) of Example 3.4 is therefore 2 4.42 4.42 2 σ 1,2 = 2 ± [( 2 ) + (2.61) ]

σ 1 = 5.63 MPa

1 2

σ 2 = −1.21 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 3.50 (CONT.) (b)

τ max =

1 2

(5.63 + 1.21) = 3.42 MPa

(c)

σ oct =

1 3

(5.63 − 1.21) = 1.47 MPa

= τ oct

1 3

2 2 [(5.63 + 1.21) 2 + (−1.21) 2 + (−5.63) = ] 2.98 MPa

______________________________________________________________________________________ SOLUTION (3.51) ( a ) We have D/d=2 and r/d=0.04. Then, we find from Figs. D.8 and D.7 that

= K b 2.6 = K t 1.9

Equation (j) of Example 3.4 is therefore 3

4(20×10 ) σ x 2.6 33.9 MPa = = π (0.125)3 3

2(5×10 ) τ xy 1.9 9.73 MPa = = π (0.125)3

Equation (i) of Example 3.4: 2 33.9 33.9 2 σ 1,2 = 2 ± [( 2 ) + (9.73) ]

σ 1 = 36.5 MPa

(b)

τ max =

1 2

(36.5 + 2.59) = 16.96 MPa

(c)

σ oct =

1 3

(36.5 − 2.59) = 11.3 MPa

= τ oct

1 3

1 2

σ 2 = −2.59 MPa

2 2 [(36.5 + 2.59) 2 + (−2.59) 2 + (−36.5) = ] 17.85 MPa

______________________________________________________________________________________ SOLUTION (3.52) We apply Eqs. (3.63). 1

2(500)(0.025×0.0375) 3 ( a ) a 0.88[ = = ] 0.635 mm (200×109 )(0.0125) 1

(b)

2 3 0.0125 p0 = 0.62[500(200 ×109 ) 2 ( 2×0.025 596.1 MPa ×0.0375 ) ] =

(c)

2 δ 1.54[(500) = ( 2(200×10 )0.0125 )] 5.339(10−3 ) mm (0.025×0.0375)

1 3

9 2

______________________________________________________________________________________ SOLUTION (3.53) ( a ) Use Eq. (3.62): 9 2

500(200×10 ) 3 = = p0 0.62[ ] 1240 MPa 4(0.025)2

( b ) Apply Eqs. (3.60) and (3.59) for r1 = r2 = r and E1 = E 2 = E to obtain the formula 1

p0 = 0.617[ PEr2 ]3 2

9 2

500(200×10 ) 3 Thus, = = p0 0.617[ ] 1959 MPa (0.025)2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.54) Using Eqs. (3.72), (3.73), and (3.74), we have

m = (1 0.4 )+4(1 0.25) = 0.6154 9

×10 ) n = 43((200 = 2.9304(1011 ) 1−0.32 )

) cos α = ± ((11 00..44 ))+−((11 00..25 25 ) = 0.2308

Interpolating Table 3.3:

ca = 1.1774

α = 76.66o

cb = 0.8616

Apply Eqs. (3.71):

4(10 )(0.6154) 3 = = a 1.1774[ ] 2.393 mm 2.9304×1011 4(10 )(0.6154) 3 = b 0.8616[ = ] 1.752 mm 2.9304×1011 3

4(10 ) Thus, = p0 1.5 = 455.5 MPa π (2.393×1.752)10−6

______________________________________________________________________________________ SOLUTION (3.55) Use Eqs. (3.67): 3

2.5(10 )(200×10 ) 2 = p0 0.418[ = ] 418 MPa 0.1(0.005) 3

)(0.005) 2 2a 2{1.52[ 2.5(10 ] } 2(0.038) = = = 0.076 mm 200×109 (0.1)

______________________________________________________________________________________ SOLUTION (3.56) Use Eqs. (3.64) and (3.65), for ν= 1 6

E2= E , r= r2= r : ν 2= 0.25, E= 1 1

2(10 )(200×10 )(2) 2 = p0 0.412[ = ] 824 MPa 0.2 6

2(10 )(0.2) 2 = 2a 2{1.545[ 200 = ] } 2(1.545) = 3.09 mm ×109 (2)

______________________________________________________________________________________ SOLUTION (3.57) Refer to Example 3.6.

θ =π 2 1 r1 ' = 0 1 r2 ' = 0 m = (1 0.5)+4(1 0.2 ) = 0.5714 9

×10 ) n = 43((210 = 2.9867(1011 ) 1−0.252 )

cos α = ± ((11 00..55))+−((11 00..22 )) = 0.4286 Table 3.3: ca = 1.3862

α = 64.62 o

cb = 0.7758 3

5×10 (0.5714) 3 = a 1.3862[ = ] 2.943 mm 2.9867×1011 5×10 (0.5714) 3 = b 0.7558[ = ] 1.604 mm 2.9867×1011 3

5(10 ) Thus, = p0 1.5 = 505.7 MPa π (2.943×1.604)10−6

______________________________________________________________________________________ 107 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (3.58) Refer to Example 3.6. We now have r1 = r2 = r. Thus, Equations (3.75) and (3.76) become

m = (1 r )+4 (1 r ) = 2r = 2(0.2) = 0.4

cos α = ((11 rr ))+−((11 rr )) = 0,

α = 90o

From Table 3.3 it can be concluded that surface of contact has a circular boundary: ca = cb = 1. 9

×10 ) n = 43((210 = 2.98667(1011 ) 1−0.252 ) 3

5(10 )(0.4) 3 a= b= 1[ 2.98667 ] = 1.885 mm ×1011

Thus,

5(10 ) = p0 1.5 = 671.9 MPa π (1.885)2 10−6

______________________________________________________________________________________ SOLUTION (3.59) Case B ( 1st column ), Table 3.2 with r1 = r2 , (a)

E1 = E2 .

m = 1r + 1r = 2r

∆ = E1 + E1 = E2

0.17 3 F r = a 0.88 = 0.88 3 400 = 0.604 mm E 210(109 )

( b= ) p0

400 P 1.5 = 1.5 π (0.604) = 524 MPa 2 π a2 (10−6 )

( c ) Equations (3.68) at z=0:

σx = σy = − p0 [(1 +ν ) − 12 ] = − 1+22ν p0 = −0.8(524) = −419 MPa σz = − p0 = −524 MPa

τ yz = τ xz =12 (σ x − σ z ) = − p2 (0.8 − 1) = 52.4 MPa 0

______________________________________________________________________________________ SOLUTION (3.60) Refer 2nd column of C, Table 3.2. Hence

E= E2= E= 210 GPa 1

∆=

2 E

2 210(109 ) 3

1 105(109 )

r= 8 mm 1 n=

1 0.008

r2= 50 mm

F= 240 kN m 1

1 − 0.05 = 105

240(10 ) ( a ) a 1.076[ = = ] 2 0.159 mm 105(109 )(105)

(b) = p0

2 F

π aL

240(103 )

= 961 MPa

π 0.159(10−3 )

______________________________________________________________________________________ 108 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (3.61) We have r1 = 0.48 m, r2 = 0.34 m, r1 ' = ∞, r2 ' = ∞, and θ = 90 . Thus, using o

Eqs.(3.72) through (3.74):

4 n = 0.796 = 1 1 + 0.48 0.34 A = m2 , B = ± 12 ( r11 − r12 )

= m

4(210×109 ) 3(1− 0.09)

= 3.0769(1011 )

0.48 −1 0.34 B cos α = 0.171, 80.15o α= ± 11 0.48 +1 0.34 = A =

1.126, = cb 0.892 . Hence

Interpolating from = Table 3.3: ca 3

1 3

4(10 )(0.796) = a 1.126[ = ] 2.4537 mm 3.0769(1011 ) 4(10 )(0.796) 3 = b 0.892[ = ] 1.9438 mm 3.0769(1011 )

Thus = p0

) F 1.5 = 1.5 π (2.45374(10 = 400 MPa π ab ×1.9438×10−6 )

______________________________________________________________________________________ SOLUTION (3.62) Given quantities are:

Thus,

r1 = r1 ' = 0.025 m r2 = −0.03 m r2 ' = ν= 0.3 −0.125 m E = 200 GPa

m =

4 0.10345 = 1 1 1 1

+ − − 0.025 0.025 0.03 0.125 9

×10 ) 9 n = 43((200 1−0.09 ) = 293.04029(10 )

Also

2 = 19.33301 A = m2 = 0.10345

= B

1 2

2 [(0) 2 + ( r12 − r1' ) 2 + 2(0)]= 12.66667 2

−1 12.66667 α cos 49.06643o = = 19.33301

From Table 3.3, we find

ca = 1.78611

cb = 0.63409

The semiaxes are then 1

1800(0.10345) 3 = a 1.78611[ 293.04029(10 = 0.00154 = m 1.54 mm 9 ] ) 1

1800(0.10345) 3 = b 0.63409[ 293.04029(10 = 0.00055 = m 0.55 mm 9 ] )

Maximum contact pressure is therefore

1800 = p0 1.5 = 1014.7 MPa π (1.54×0.55)(10−6 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (3.63) Now given data is as follows:

r1 = r1 ' = 0.02 m r2 = −0.022 m r2 ' = −0.125 m ν= 0.3 E = 200 GPa

Therefore,

= 0.08594

4 1 1 1 1 + − − 0.02 0.02 0.022 0.125 9

9 ×10 ) n = 43((200 1−0.09 ) = 293.0403(10 )

We have

A = m2 = 23.2721 1 1 B = ± 12 [ − 0.022 + 0.125 ] = 18.7273 o .7273 α = cos −1 18 23.2721 = 36.42

Using Table 3.3

ca = 2.323

cb = 0.541

Then, the semiaxes are: 1

= = = a 2.323[ 1800(0.08594) ]3 0.00188 m 1.88 mm 293.0403×109 1

= b 0.541[ 1800(0.08594) = ]3 0.00044 = m 0.44 mm 293.0403×109 Maximum contact stress is now obtained as

1800 p0 1.5 1039 MPa = = π (1.88×0.44)10−6

End of Chapter 3

CHAPTER 4 SOLUTION (4.1)

σ1 = 0

60 mm

σ2

M=1.5 kN ⋅ m 3

P Mc 4P 32(1.5 ×10 ) = − − 2 A I π (0.06) π (0.06)3 = −353.7 P − 70.74 ×106

− − σ2 =

We have

σ 12 − σ 1σ 2 + σ 22 = σ yp2 So,

σ 2 = σ yp

353.7 P + 70.74 ×106 = 250 ×106 ,

Pall = 507 kN

______________________________________________________________________________________ SOLUTION (4.2)

= p 7.86(9.81)( = π d 2 4) 60.6d 2 kN m p=60.6d2 T=325 N ⋅ m

2.5 m

σx =

32 M πd3

τ = 16T π d 3

32( pL2 8) 32(60.6d 2 )(25)103 1.93 ×106 = = 8π d 3 d πd3 3 16(325) 1.66 ×10 − = − τ= 3 d3 πd

= σx

Equation (4.9a), σ x + 3τ 2

2 = σ all

1.93 ×106 2 −1.66 ×103 2 ( ) + 3( ) = (100 ×106 ) 2 3 d d

3.725 ×106 8.267 + 6 = 100 ×108 2 d d Solving by trial & error: d = 32.8 mm. Use a 33 − mm diameter shaft. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.3)

σy = 100 MPa τ xy = −80 MPa σx = 0

σ yp = 250 MPa (Table D.1) Hence

σ yp −σ y 2 ( ) + τ xy2 σ 1,2 =± 2

= 50 ± (−50) 2 + (−80) 2 = 50 ± 94.3 or (a)

σ 1 = 144.3 MPa

Maximum shear stress memory:

σ1 − σ 2 ≤

σ 2 = −44.3 MPa

σ yp n

144.3 − (−44.3) ≤ 188.6 > 166.7

250 1.5

Failure will occur. (b)

Maximum energy of distortion theory:

σ 12 − σ 1σ 2 + σ 22 ≤ (

σ yp n

)

(144.3) 2 − (144.3)(−44.3) − (−44.3) 2

≤ 166.7

158.9 < 166.7

Failure will not occur.

______________________________________________________________________________________ SOLUTION (4.4) State of stress is given by

σ 1 = σ = π32( 0.M1) + π (40P.1) , 3

Refer to Table 4.1 and Eq. (2.66): 0.47σ yp = 0.47σ or Thus, Solving,

σ2 = σ3 = 0 σ yp = σ

) 221(10 3 ) = π32( 0(17 + π (40P.1)2 .1)3

P = 375.7 kN

______________________________________________________________________________________ SOLUTION (4.5) From Table D.1: σ yp = 250 MPa We have

σ 1, 2 = −302+90 ± [( −302−90 ) 2 + ( −40) 2 ]

1 2

(CONT.) ______________________________________________________________________________________ 112 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 4.5 (CONT.)

σ 1 = 102.1 MPa

σ 2 = −42.1 MPa

( a ) Equation (4.2a) gives

102.1 + 42.1 = 250 n n = 1.73

( b ) Equation (4.5a) leads to

2 (102.1) 2 − (102.1)( −42.1) + ( −42.1) 2 = ( 250 n )

from which

n = 1.95

______________________________________________________________________________________ SOLUTION (4.6) Referring to Appendix B, we obtain

σ 1 = 101.3 MPa

σ 3 = −51.32 MPa

σ 1 − σ 3 = σ yp

(a)

σ yp = 101.3 + 51.32 =152.6 MPa

or (b)

σ2 = 0

τ= oct

Thus,

1 3

[(101.3) 2 + (101.3 + 51.32) 2 + (51.32) 2= ] 2 63.41 MPa

= σ yp 63.41 = 0.47 134.9 MPa

______________________________________________________________________________________ SOLUTION (4.7) Using Eq. (4.6), we have 1

Here

0.47 nyp = 13 [(σ 1 − σ 2 ) 2 + σ 12 + σ 22 ] 2

σ 1, 2 = σ2 ± 12 [σ 2 + 4τ 2 ] and

σ = 32πdM + π4dP , 3

(a)

1 2

T τ = 16 πd

Substituting the given data: 3

σ 1, 2 = 12 [ 32π ((04.×1210) ) + 4π((450.×1210) ) ] 3

± 12 {[ 32π ((04.×1210)3) + 4π((450.×1210)2) ]2 + 4[ 16π(11( 0..212×10)3 ) ]2 } 2 or

σ 1 = 49.55 MPa

Equation (a) is then

σ 2 = −21.99 MPa 1

Solving,

2 2 2 2 1.41 280 n = [( 71.54) + ( 49.55) + ( −21.99) ]

n = 4.4

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.8) Maximum stresses, occurring at the fixed end are:

σ = McI = 4502(t0.253 t ) = 168t .75 4

= 337t .5 τ = 23 450 2t 2

From Eq. (4.9a), we have

2 σ yp = σ 2 + 3τ 2 = ( 280 × 10 6 ) 2 Therefore, at neutral axis σ = 0 : σ yp = 3τ gives t = 1.45 mm At the extreme fibers, τ = 0 : σ yp = σ gives t = 8.45 mm

Allowable width is thus

tall = 8.45 mm

______________________________________________________________________________________ SOLUTION (4.9) We have

σ yp

τ yp =2 = 175 MPa, σ 1 = −σ 2 = τ 175 1.5

(a) or

) = 16π(d500 3

d = 27.95 mm

σ 1, 2 = σ2 ± 12 σ 2 + 4τ 2

(b) Here

σ = 32πdM = 32 (πpLd 8) 3

= 8π32d 3 [ 14 πd 2 (77 × 10 3 )10 2 ] = 77(105 ) d

= τ

16T

π d3

16(500)

= 2546.48 d 3

π d3

Using Eq. (4.8a): σ yp n

= σ 2 + 4τ 2 6

350(10 ) = [( 77×d10 ) 2 + 4( 2546.48 )2 ]2 1.5 d3 5

5.444(1016 ) = 5.929d(210 ) + 2.594d (610 ) Solving, by trial and error:

= d 0.0368 = m 36.8 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.10)

(a) (b)

We have= σ all

90 = 1.2 75 MPa

σ all = σ 1 − σ 3 = 63.4 + 12.2 = 75.6 > 75

Failure occurs

2 2σ all = (63.4 − 0.53) 2 + (0.53 + 12.2) 2 + ( −12.2 − 63.4) 2

σ all = 70 < 75

No failure

______________________________________________________________________________________ SOLUTION (4.11) ( a ) Using the torsion formula,

τ = π T( 0(.005.05) ) 2 = 5093T 4

and σ 1 = −σ 2 = 5093T Equation (4.5a) yields then

[280(10 6 )]2 = (5093T ) 2 + (5093T )(5093T ) + ( −5093T ) 2

Solving,

= T 31.74 kN ⋅ m

( b ) We now have

= σ

400(103 )π

= 160 MPa

π (0.05)2

τ = 5093T

(as before)

Principal stresses are: 6

σ 1, 2 = 160(210 ) ± 12 [(160 × 10 6 ) 2 + 4(5093T ) 2 ] 

1 2



=a±b With this notation, Eq. (4.5a) becomes

2 = ( a + b) 2 − ( a 2 − b 2 ) + ( a − b) 2 = a 2 + 3b 2 σ yp

Thus, Solving,

( 280 × 10 6 ) 2 = ( 160×210 ) 2 + 43 [(160 × 10 6 ) 2 + 4(5093T ) 2 ] 6

= T 26.05 kN ⋅ m

______________________________________________________________________________________ SOLUTION (4.12) Maximum moment is

M = and hence

PL2 8

σ= σ= 1

6(1.5)2 8

= 1.688 kN ⋅ m

My I

1.688(0.125)103

= 1.62 MPa

0.1(0.25)3 12

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.12 (CONT.) (a)

σ yp n

= σ 12 − 0 = σ 1

28 n

= 1.62

from which

n = 17.3 σ yp

σ1 − 0 = n

(b)

or n = 17.3 ______________________________________________________________________________________ SOLUTION (4.13) Table D.1; σ yp = 250 MPa

= σx

, = τ xy

16T

π d2

π d3

Equation (4.8a) results in

τ xy=

1 2

π d3

2 [( nyp ) 2 − σ x2 ]=

1 2

45×10 2 2 {( 2501.6×10 ) 2 − [ π4×(0.054) 77.5 MPa 2 ] }= 6

Thus 16

τ xy =

π (0.054)3 16

(77.5 ×106 )= 2.396 kN ⋅ m

______________________________________________________________________________________ SOLUTION (4.14) Refer to solution of Prob.4.13, Equation (4.9a) gives

= τ xy

1 3

2 [( nyp ) 2 − σ= x]

1 3

45×10 2 2 {( 2501.6×10 ) 2 − [ π4×(0.054) } 89.5 MPa 2 ]= 6

and

π d3 16

τ xy =

π (0.054)3 16

(89.5 ×106 )= 2.767 kN ⋅ m

______________________________________________________________________________________ SOLUTION (4.15) Referring to Appendix B, we compute

−1.521 MPa −4.528 MPa σ1 = 12.05 MPa σ2 = σ3 =

Using Eq. (4.5a),

Solving,

2 2σ yp = (12.05 + 1.521) 2 + ( −1.521 + 4.528) 2 + ( −4.528 − 12.05) 2 = 468

σ yp = 15.3 MPa

Hence,

= τ yp 15.3(0.577) = 8.828 MPa Therefore

9 = σ y 2(= 2.039 MPa 8.828 )

9 = σ x 3(= 3.058 MPa 8.828 )

______________________________________________________________________________________ 116 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (4.16) y A

P T

1.2 m

σx

τ We have P = 50 R, T = 0.8 R, and M = 1.2 R . Stresses are (1.2 R ) σ b = 32πdM = 32 = 97,784.8R π ( 0.05 ) 3

T τ = − 16 = πd 3

−16 ( 0.8 R ) π ( 0.05 )3

= −32,595R

σ a = π ( 050.05R) 4 = 25,464.8R 2

σ x = σ a + σ b = 123,249.6 R ( a ) Equation (4.8a): 1

260 (106 ) 2

= R[(123,249.6) 2 + 4( −32,595) 2 ] 2 R = 932 N

( b ) Equation (4.9a): 1

130 × 10 6 = R[(123,249.6) 2 + 3( −32,595) 2 ] 2 R = 959 N

or ______________________________________________________________________________________ SOLUTION (4.17) Referring to Appendix B,

σ1 = 197.4 MPa σ2 = σ3 = −14.44 MPa −72.96 MPa

Applying Eq. (4.4b):

2 2σ yp = (197.4 + 14.44) 2 + ( −14.44 + 72.96) 2 + ( −72.96 − 197.4) 2

σ yp = 246.4 MPa

Hence, τ yp 246.4(0.577) = = 142.2 MPa Thus, = σy

140 = 40 39.38 MPa 142.2

140 = σ x 50 = 49.23 MPa 142.2

______________________________________________________________________________________ SOLUTION (4.18) Stresses are

.25 ) σ 1 = prt = p0(.0005 = 50 p

σ 2 = 2prt = 25 p

(CONT.) ______________________________________________________________________________________ 117 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ (CONT.) ( a ) Applying Eq. (4.5a),

(50 p ) 2 − 1250 p 2 + ( 25 p ) 2 = ( 280) 2 p = 6.466 MPa

or ( b ) Using Eq. (4.2a),

50 p − 0 = 280

p = 5.6 MPa

______________________________________________________________________________________ SOLUTION (4.19) We have (σ yp ) all=

= 68.33 MPa .

82 1.2

Referring to Appendix B:

σ 1 = 63.44 MPa

σ 2 = 0.533 MPa

σ 3 = −12.17 MPa

( a ) Using Eq. (4.1),

σ yp = 63.44 + 12.17 = 75.6 > 68.33 :

Failure occurs

( b ) Applying Eq. (4.4b)

2 2σ yp = (63.44 − 0.533) 2 + (0.533 + 12.17) 2 + ( −12.17 − 63.44) 2

σ yp = 70.13 > 68.33 :

Failure occurs

______________________________________________________________________________________ SOLUTION (4.20) Referring to Appendix B:

= σ 1 162.4 = MPa σ 2 46.15 = MPa σ 3 1.468 MPa ( a ) Equation (4.1) yields

n = 162.4300 −1.468 = 1.86

( b ) Equation (4.4b) gives

2σ 2

) n 2 = (σ −σ )2 +(σ −σyp )2 +(σ −σ )2 = (116.25)2 +( 442 (.300 682 ) 2 + (160.932 ) 2 1

or n = 2.08 ______________________________________________________________________________________ SOLUTION (4.21) Referring to Appendix B:

= σ 1 156.2 = MPa σ 2 42.13 = MPa σ 3 11.7 MPa ( a ) Equation (4.1) gives

n = 156.220 2 −11.7 = 1.52

( b ) Equation (4.4b) yields

2σ 2

) n 2 = (σ −σ )2 +(σ −σyp )2 +(σ −σ )2 = (114.07 )2 +(230( 220 .43) 2 + ( −144.5 ) 2

n = 1.67

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.22) We have σ1 = τ ,

σ 2 = −τ . Table D.1: σ u 170 = = σ u' 650 MPa MPa, (a)

Equation (4.12a) is thus

−τ 1 − 650 = 1.6 τ = 84.2 MPa

170

(b)

or Equation (4.11a):

τ=

σu

= 106.3 MPa

170 1.6

______________________________________________________________________________________ SOLUTION (4.23)

σθ

τ Tc J

J ≈ 2π r 3t pr 5(106 )(105) = σ= = 52.5 MPa θ t 10 pr σ= = 26.25 MPa a 2t

σa

50 ×103 (0.21) = −144.4 MPa 2π (0.105)3 (0.01)

− = − τ=

26.25 + 52.5 26.25 − 52.5 2 2 ± ( ) + (−144.4) = 39.4 ± 145 2 2 σ 1 = 184.4 MPa σ 2 = −105.6 MPa σu 250 Thus, ;= 184.4 = , n 1.4 = σ1 n n σu ' −520 and ; , − 105.6 = = n 4.9 σ= 2 n n

= σ 1,2

______________________________________________________________________________________ SOLUTION (4.24)

M 0 : 150(2)(1) − R (1.2) = 0, ∑= = : R 50 kN ∑ F 0= A

RB = 250 kN

150 kN/m A 50 kN V, kN

1.2 m

0.8 m

C b

250 kN 120

x -130

(CONT.) ______________________________________________________________________________________ 119 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 4.24 (CONT.)

3V 3 130 ×103 97.5 ×103 = = = τ max 2 A 2 2b 2 b2 τ max = σ 1 = −σ 2

And Thus,

τ max = σ all ;

97.5 ×103 = 120 ×106 , b2

b= 28.5 mm

Use a 30 mm by 60 mm rectangular beam. ______________________________________________________________________________________ SOLUTION (4.25)

-420

σ 2 = −4σ 1

σ1

260

( a ) Upon following the procedure described in Sec. 4.11, Mohr’s circle is constructed as shown in the sketch above. The circle representing the given loading is then drawn by a trial and error procedure, as is indicated by the dashed lines. From the diagram, we measure the following values:

σ 1 = 77 MPa

σ 2 = −308 MPa

( b ) Applying Eq. (4.12a), or Solving,

σ1 σu

− σσu2' = 1

σ1

4σ 1 − −420 =1

260

σ 1 = 75 MPa

σ 2 = −300 MPa

______________________________________________________________________________________ SOLUTION (4.26) Principal stresses are 2 180 2 σ 1, 2 = −180 2 ± [( 2 ) + 200 ]

σ 1 = 129.3 MPa,

1 2

σ 2 = −309.3 MPa

( a ) Equation (4.11a):

σ 1 < 290 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.26 (CONT.) But since

σ 2 > 290 MPa :

failure occurs

( b ) Equation (4.12a):

.3 − −309 650 = 1 gives 0.446 + 0.476 = 0.922 < 1 129.3 290

Thus, no fracture Note that Coulomb-Mohr theory is the most reliable when σ u ' >> σ u , as in this example.

______________________________________________________________________________________ SOLUTION (4.27)

σ x = 2prt + 2πPrt 6

2.8(10 )125 45(10 ) = + 2π (0.125)(0.005) = 46.46 MPa 2(5) pr t

σ= y = τ

Tr 2 π r 3t

2.8(106 )125 5

= 70 MPa

31.36(103 )

= 63.89 MPa

2π (0.1252 )(0.005)

Thus,

σ 1, 2 = 12 ( 46.46 + 70) ± [ 14 ( 46.46 − 70) 2 + (63.89) 2 ] or

σ 1 = 123.2 MPa

1 2

σ 2 = −6.74 MPa

( a ) Equation (4.12a),

6.74 − −500 =1 gives 0.587 + 0.013 = 0.6 < 1 123.2 210

Thus,

no fracture

( b ) Equations (4.11a) shows

123.2 < 210, 6.74 < 210,

no fracture

no fracture ______________________________________________________________________________________ SOLUTION (4.28) State of stress is represented by Mohr’s circle shown below.

τ τ

O 3 4

σu

3 8

σu ( 43 σ u , τ ) C

2θ p '

5 8

σu

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.28 (CONT.) From the circle, we obtain (a)

τ = σ u ( 85 ) 2 − ( 83 ) 2 = 12 σ u

(b)

θ p ' = 12 tan −1 13 28 = 26.57 o Orientation of the fracture plane is shown below. Fracture surface

T θp' P P ______________________________________________________________________________________ SOLUTION (4.29) Principal stresses are

σ 1, 2 = 12 ( 200 + 20) ± [(90) 2 + (150 2 )] or

σ 1 = 284.9 MPa 284.9 = 420 n , 420 64.9 = n ,

(a)

1 2

σ 2 = −64.9 MPa

n = 1.47 n = 6.47

−64.9 .9 1 ( b ) Equation (4.12a): 284 420 − 900 = n

Solving, n = 1.33 ______________________________________________________________________________________ SOLUTION (4.30) Uniform shear stress τ acts on a typical element as shown.

τ = P πdt σ 1 = −σ 3 = τ

σ1 = σ u

(a)

πtd

(b)

σ1 σu

= σu,

− σσu3' = 1;

σ3 = σu

P = πtdσ u

τ (1 + σσ ' ) = σ u u

P = (1+σtduσ uσ u ')

______________________________________________________________________________________ SOLUTION (4.31) Table 4.3: K c 23 = = 1000 MPa mm σ yp 444 MPa Note that the values of crack length a and plate thickness t satisfy Table 4.3. 25 Table 4.2: wa = 125 = 0.2 λ = 1.37

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.31(CONT.) Equation (4.18), with n=1:

σ= all

Kc =

= 59.90 MPa

23 1000 (1.37) π (25)

λ πa

Therefore

= = = 187.2 kN P σ all ( wt ) 59.90(125 × 25) all

Note, the nominal stress at fracture:

= σ

187.2(103 )

= 93.6 MPa < 444 MPa

P t ( w− a )

20×10−3 (125 − 25)×10−3

______________________________________________________________________________________ SOLUTION (4.32) We have a w = 0.005 5 = 0.001 and λ = 1 by Table 4.2. From Table 4.3:

MPa m K c 59 = = σ yp 1503 MPa ( a ) K λσ = = π a (1)(100) π= (0.03) 30.7 MPa m

n = KKc = 3059.7 = 1.92 ( b ) Using Eq. (4.18) with n = 1 :

= σ

Kc =

λ πa

= 192.2 MPa

59 (1) π (0.03)

This is well below the yield strength. ______________________________________________________________________________________ SOLUTION (4.33) By Table 4.3: K c = 66 We have

a w

1000 MPa mm and σ yp = 1149 MPa. Table 4.2:

= 10 65 = 0.15

= σ

Kc =

λn π a

λ = 1.02 = 165.9 MPa

66 1000 (1.02)(2.2) π (10)

Thus

t= req

P 2 wσ

200(103 ) 2(65)(165.9)

= 9.27 mm

A thickness of 9.3 mm should be used. Note that both values of a and t satisfy Table 4.3. ______________________________________________________________________________________ SOLUTION (4.34) 8 Refer to Example 4.5. We now have wa = 40 = 0.2 .

Table 4.2:

λa = 1.37

λb = 1.06

Table 4.3: = K c 59 = 1000 MPa mm σ yp 1503 MPa Equation (a) of Example 4.5, with M = 0.17 P : P) λσ = (1.37) ( 0.04 )(P 0.01) + 1.06 ( 06.01( 0)(.170.04 = 71,000 P )2 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.34 (CONT.) By Eq. (4.18): 6

λσ = n Kπa ;

1000 (10 ) 71,000 P = 59 1.8 π ( 0.008 )

from which

P = 92.09 kN

The nominal stress at fracture: 3

σ = t ( wP−a ) = ( 0.0192)(.009.04(10−0.)008) = 287.8 MPa <1503 MPa ______________________________________________________________________________________ SOLUTION (4.35) From Table 4.3: K c = 23 1000 MPa

mm and σ yp = 444 MPa

. λ = 112

Case B of Table 4.2: a w = 016 . , By Eq.(4.18), with n = 1 :

= σ

Kc =

λ πa

= 81.93 MPa

23 1000 1.12 π (20)

It follows that

P= σ ( wt= ) 81.93(150 × 30)= 369 kN

Then

σ =

P ( w− a ) t

369(103 ) (0.15 − 0.03)(0.03)

= 102.5 MPa

= 102.5 MPa < σ yp ______________________________________________________________________________________ SOLUTION (4.36) Case A of Table 4.2 and Table 4.3:

σ yp = 1503 MPa

K c = 59 1000 MPa mm λ = 1.01 (assumed)

By Eq.(4.18):

= σ

Kc =

nλ π a

(a)

σ=

pf r t

(b)

σ =

pf r 2t

= 213 MPa < σ yp

59 1000 (2)(1.01) π (6)

p= f

σt r

213(5) 30

= 35.5 MPa

p f 2(35.5) = = 71 MPa

______________________________________________________________________________________ SOLUTION (4.37)

= = MPa m σ yp 392 MPa Table 4.3: K c 31 a Case D of Table 4.2:= w

0.4

∴ = λ 1.32

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.37 (CONT.) Using Eq.(4.18) with n = 1 and σ = 6 M tw : 2

6M 6M K c λ= = π a; 31(106 ) 1.32 0.03(0.12) π (0.048) 2 tw2

Solving

= M 4.35 kN ⋅ m

______________________________________________________________________________________ SOLUTION (4.38)

= = MPa m and σ yp 1503 MPa (Table 4.3). ( a ) K c 59 Case B of Table 4.2: with a=12 mm and w=125 mm: wa = 01 .

Eq.(4.18) with n = 1 :

∴ λ = 112 .

= 59 (1.12) = σ π (0.012), σ 271.3 MPa Therefore

Pall = ( wt )σ = (125 × 25)(271.3) = 848 kN

Note that the nominal stress at fracture

= σ

848(10 ) = 300.2 MPa < σ yp

25(125 −12)(10−6 )

( b ) Table 4.3: K c 66 = = MPa m and σ yp 1149 MPa Thus Solving

×10 66 σ π a (1.12)( 125848 ) πa = ×106 (1.12)= ×25×10−6 3

= a 0.015 = m 15 mm

Observe this value of a satisfies Table 4.3 ______________________________________________________________________________________ SOLUTION (4.39) The state of stress on the cylinder wall is considered to be biaxial. Maximum principal stress, that is, tangential stress, in the cylinder has the mean and range values pm r t

= σm

where

= , σa

pa r t

( pmax + pmin ) = ( pmax − pmin ) =

pm = pa =

1 2 1 2

σa σm

pa pm

(1) 1 2 1 2

(4.2 + 1.0) = 2.6 MPa (4.2 − 1.0) = 1.6 MPa

Since stresses are proportional to pressures, we have

1.6 2.6

Substitution of the data into Eq.(c) of Sec. 4.15.3:

σu n 400 2.2 = = 71.6 MPa 1.6 400 + 1 2.6 160 + 1

= σm

σa σu σm σe

Using Equation (1), we have

= t

pm r

σm

2.6(800) 71.6

= 29.1 mm

This is the minimum safe thickness for the pressure vessel. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.40)

= σm

120(103 )

= 48 MPa

25(10−4 )

and

σa σ cr

+ σσ mf = 1;

FA = 186.3 kN

1200 FA

240 (106 )

48 (10 ) + 700 =1 (106 )

______________________________________________________________________________________ SOLUTION (4.41)

σ 1a − σ 3a = 15 p − 0 = σ ea σ 1m − σ 3m = 9 p − 0 = σ em

Then,

15 p 250 (106 )

p + 3009(10 =1 6 )

p = 11.11 MPa

______________________________________________________________________________________ SOLUTION (4.42) We have

σ cr = 1−(σσ σ )

(a)

where,

σ m = F 2+AF , max

min

σ a = F 2−AF max

min

(b)

Substituting Eqs. (b) into (a):

σ cr = 1−[ ((FF +−FF )) 22AAσ ]

Solving,

max

min

max

min

A = σ1cr [ Fmax − 12 ( Fmax + Fmin )(1 − σσcru )] ______________________________________________________________________________________ SOLUTION (4.43) We have σ m = σ a . Using Table 4.4: σm

510 1.5

σm + 1050 1.5 = 1

from which σ m = 228.8 MPa. At the fixed end:

M max = PL = 10(0.05) = 0.5 N ⋅ m

Hence,

= M a ,m

( M max ± M min ) 2

= 0.25 N ⋅ m

and 0.25 ) = 300 = 228.8(10 6 ) σ a = σ m = 6btM = 06.(005 t t m 2

Solving,

t = 1.145 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.44) We have σ a = σ m . From Table 4.4: σm

740 2.5

σm + 1500 2.5 = 1

σ m = 198.2 MPa.

At the center of the beam:

M = PL = 4 (0.25)(20)(0.125) = 0.625 N ⋅ m max

Hence,

( M max ± M min ) 2

= 0.3125 N ⋅ m

= M a ,m and

) σ a = σ m = 6btM = 6(00..013125 = 187t .5 = 198.2(10 6 ) t m 2

Solving,

t = 0.973 mm

______________________________________________________________________________________ SOLUTION (4.45)

σ max = σ min = McI , and

σ m = 4πMr , 3

τ max = TrJ ,

τ min = 0,

σx =σ,

τ xy = τ ,

σa = 0

τ m = τ a = πTrr

σ y = σ z = τ xz = τ yz = 0

Equations (4.21) yield

0 + 3τ a2 = σ ea ,

σ m2 + 3τ m2 = σ em

Then, Soderberg relation becomes

σ cr = or

3τ a

σ m2 + 3τ m2 1− σ yp

σ cr = σσ

cr yp

σ m2 + 3τ m2 + 3 τ a

or 1 2

σ cr = σσ [( 4πMr ) 2 + 3( πTr ) 2 ] + π3rT cr

Solving this expression for r, we obtain Eq. (P4.45). ______________________________________________________________________________________ SOLUTION (4.46)

σ 1a , 2 a =

σ xa +σ ya 2

σ xa

σ −σ

2 ]2 ± [( xa 2 ya ) 2 + τ xya 2

2 ] = 2 ± [ σ4xa + τ xya 4

81(10 ) = 900 + 4(10 4 )] 2 2 ±[ 4

σ 1a = 942.44 MPa,

σ 2 a = −42.44 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.46 (CONT.) Similarly, Thus,

σ 1m = 161.8 MPa, σ 2 m = −61.8 MPa σ ea = σ 1a − σ 2 a = 984.88 MPa σ em =σ 1m + σ 2 m = 223.6 MPa

( a ) Modified Goodman relation:

= σ cr

= 1086 MPa

984.88 1− (223.6 2400)

×2400 800 ) b = ln( 0ln(.910 = −0.0863 3 108 ) −11.587 = 2.89(106 ) cycles N cr = 103 ( 0.91086 ×2400 )

( b ) Soderberg criterion:

= σ cr

= 1145 MPa

984.88 1− (223.6 1600)

−11.587 3 = = N cr 10 ( 0.91145 1.56(106 ) cycles ×2400 )

( c ) The SAE criterion:

σ cr = 1086 MPa, N cr = 1(

( d ) Gerber criterion:

= σ cr

b = −0.0596

1086 −16.778 2400

= 0.59(106 ) cycles

)

= 991.32 MPa

984.88 1− (223.6 2400)2

3 991.32 −11.587 = = N cr 10 ( 0.9×2400 ) 8.30(106 ) cycles

______________________________________________________________________________________ SOLUTION (4.47)

= I

25(4)3 12

= 133 mm 4 .

bh3 12

For a wide cantilever beam (see Table 3.1, and Table D.4): 3

δ = (1 − ν 2 ) 3PLEI = 0.91 3PLEI

Thus

3(210×109 )(133×10−12 )

= δ min

This gives = Pmin

3 EI 0.91L3

0.91(0.26)3

(0.012)

= 62.87 N and hence Pmax = 125.74 N

= Pm 94.3 = N, Pa 31.4 N and = σm

94.3(0.26)(2×10−3 )

= 368.7 MPa 0.133(10−9 )

31.4 = σ a 368.7( = 122.8 MPa 94.3 )

Equation (c) of Section 4.15.3:

= n

σu = σu

σm +

σa

368.7 +

= 1.45

860 860 (122.8) 470

______________________________________________________________________________________ 128 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (4.48) Use Eq.(e) of Sec.4.15.3: σ

= n

yp = σ yp

σm +

σe

σa

368.7 +

= 1.14

600 600 (122.8) 470

______________________________________________________________________________________ SOLUTION (4.49)

σ xa = (800 + 600) 2 = 700 MPa (800 − 600) 2 = 100 MPa σ xm = σ ya = (500 + 300) 2 = 400 MPa σ ym = (500 − 300) 2 = 100 MPa (200 + 150) 2 = 175 MPa τ xya = (200 − 150) 2 = 25 MPa τ xym = Equations (4.21) give then

2σ ea2 = (700 − 400) 2 + 400 2 + 700 2 + 6(175) 2 2 2σ em = (100 − 100) 2 + 1002 + 1002 + 6(25) 2

or = σ ea 679.61 = MPa, σ em 108.97 MPa ( a ) Modified Goodman criterion:

σ cr =

= 729.28 MPa

679.61 1− (108.97 1600)

×1600 0.5×1600) = −0.051 b = ln(0.9ln(10 3 108 ) 3 729.28 −11.752 = N cr 10 = ( 0.9×1600 ) 2.97(106 ) cycles

( b ) Soderberg criterion:

σ cr =

= 762.72 MPa

679.61 1− (108.97 1000)

3 762.78 −11.752 ( 0.9×1600 ) 1.75(106 ) cycles N cr 10 = =

______________________________________________________________________________________ SOLUTION (4.50)

σ a = (M σm =

max − M min ) c

( M max + M min ) c 2I

M = PL

Substituting these into Soderberg relation, we obtain

M max = (21σ+ λcr)Ic + 11+−λλ M min where

λ = σσ = 23

Thus,

13 200×10 )( 0.05×0.1 ) Pmax = c (21σ+crλ I) L + 11+−λλ Pmin = 2 (12 ( 5 3)( 0.05 )(1.2 ) + 5 3 (10,000)

= 16, 667 + 2000 = 18.7 kN ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (4.51) We have W = mg = 80 × 9.81 = 784.8 N From Eq. (4.29);

σ max = (1 + 1 + δ2 h ) WA st

Solving, with δ st = WL AE , we obtain max h = L2σEW ( Aσ max − 2W )

(a)

Substituting the given data:

= h

(2)(350×106 ) 2(105×109 )(784.8)

= − 1569.6) 0.365 m (250 × 350

______________________________________________________________________________________ SOLUTION (4.52) W b=40 mm

h L

d=80 mm

W R We have

= I

v= st Thus,

1 12

δP

δ st =

R k

3 (0.04)(0.08)= 1.7 ×10−6 m 4

WL3 3 EI

50(1.6)3 3(200×109 )(1.7×10

= 0.2 mm

−6

)

(W − R) L3 RL3 R vst = =0.2 ×10−3 − = 3EI 3EI k 1 L3 1 (1.6)3 200 ×10−6 = R( + ) = R( ) + k 3EI 240 ×103 3 × 200 ×103 ×1.7 R = 24.4 N and W − R = 25.6 N

= 200 R(4.1667 + 4.0157) ,

Then

25.6 0.11×10−3 m = 0.11 mm = 240 ×103 25.6(1.6)(0.04) 0.964 MPa = σ st = 1.7 ×10−6

δ st =

h=300 mm.

(CONT.) ______________________________________________________________________________________ 130 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 4.52 (CONT.)

2 ( 0.3) =74.9 0.11×10−3 = vmax 74.9(0.11) = 8.2 mm K =+ 1 1+

= σ max 74.9(0.964) = 72.2 MPa ______________________________________________________________________________________ SOLUTION (4.53)

(22) 2 380 mm 2 = 4

ABC =

= AAB

Since segment BC has smaller cross section: So,

= (30) 2 707 mm 2 4

= = 79.8 kN Pmax σ= 210(380) all ABC

WL WL ) AB + ( ) BC AE AE 250 ×106 1 2 = [ = + ] 0.024 mm 9 70 ×10 707 380 2h 79,800 K =+ 1 1+ = =319.2 250 δ st = δ st (

1+ Solving

2h = 319.2 0.024

= h 1223 = mm 1.2 m

______________________________________________________________________________________ SOLUTION (4.54) Through the use of Eq. (4.29):

A [ σ max − 1]2 = 1 + 2hδ st W

σ max = WA [1 + 1 + δ2 h ], st

from which 2 A2 σ max

hAE − 2σWmax A + 1 = 1 + 2WL

Solving for W, we obtain Eq. (P4.54). ______________________________________________________________________________________ SOLUTION (4.55) The kinetic energy is

Ek = But

W ω 2r 2 2g

E k = 12 Tφ

1090(240×2π 60)2 (0.35)2 2(9.81)

= 4300 N ⋅ m

where T =

GJφ L

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 4.55 (CONT.) Thus,

φ = 2 LEk JG

(a)

Substituting the data given: 1 2

)2 φ = [ 80.52(×101.5)(π4300 ] = 0.08309 rad = 4.76o ( 0.0625 ) 9

Introducing

τL TL φ = GJ = τrJ GJL = rG

into Eq. (a), we obtain

τ = 4GE k AL

where A = πc

Substituting the numerical values, 9

4(80.5×10 )4300 2 τ [= ] 274.3 MPa = π (0.0625)2 (1.5)

______________________________________________________________________________________ SOLUTION (4.56) W 0.75 m 1.2 m We have σ yp = Mc I = PLc I from which

= P

σ yp I

= Lc

280(106 )(0.054 12) 1.2(0.025)

= 4.86 kN

End deflection is

= δ max But Solving,

4860(1.2) (12) = 26.87 mm

3(200×109 )(0.05)4

1 W (0.75 + 0.02687) = 2 (4860)(0.02687)

W = 84.05 N End of Chapter 4

CHAPTER 5 SOLUTION (5.1) Y 15 mm

A1 z

N.A. 150 mm

15 mm Z A

135 mm A1 z1 + A2 z2 A1 + A2

= z

(150×15)7.5 + (135×15)[15 + (135 2)] 150×15 +135×15

z= y= 43 mm

or Then,

I y = 121 (150)(15) 3 + (150 × 15)(35.5) 2 + 121 (15)(135) 3 + (135 × 15)(39.5) 2

9.11(106 ) mm 4 I= I= y z

I yz = (150 × 15)( −32)( −35.5) + (135 × 15)(35.5)(39.5)

= 5.4(106 ) mm 4 We have the moment components:

My = 0, Mz = −11.25(0.9) = −10.125 kN ⋅ m

Thus,

+10125(9.11)(0.043) (σ x ) A = −10125(5.4)(0.107) = −35 MPa [(9.11)2 − (5.4)2 ](10−6 )

Equation (5.15) gives

tan φ = 95..114 = 0.593

φ = 30.67o

______________________________________________________________________________________ SOLUTION (5.2) 80 mm

N.A 150 mm

A z

φ = 63.8o 1 kN 30o y

I= y

1 12

3 hb=

1 12

3 (150)(80)= 6.4 ×106 mm 4

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.2 (CONT.)

= Iz

1 12

3 bh =

1 12

3 (80)(150) = 22.5 ×106 mm 4

= M y ( P sin= α ) L 600 N ⋅ m = M z ( P= cos α ) L 1, 039.2 N ⋅ m yd = −75 mm zd = 40 mm ( a ) Equation (5.15) with I yz = 0 : M

tan φ = II zy M yz = II zy tan α

∴ φ = 63.8o

( b ) Thus, maximum tensile stress is at point A. Equation (5.16) gives

600(0.04) −0.075) σ max = 13.24 MPa − 1,039.2( = 22.5×10 6.4×10 −6

−6

______________________________________________________________________________________ SOLUTION (5.3) 3 hb =

1 12

3 (40)(100) = 3.33 ×106 mm 4

= bh3

1 12

3 (100)(40) = 0.533 ×106 mm 4

= Iy

1 12

= Iz

1 12

= M y M o cos = α 800(cos= 25o ) 725 N ⋅ m 800(sin 25o ) 338.1 N ⋅ m = M z M= = o sin α ( a ) Equation (5.15) with I yz = 0 : M

tan φ = II zy M yz = II zy cot α

∴ φ = 18.9 o

( b ) There, maximum compressive stress is at A. 100 mm 40 mm

N.A A

Equation (5.16) results in

25o y

φ = 18.9 o 800 N ⋅ m

725( −0.05) 338.1(0.02) σ max = − 0.533(10 = −23.57 MPa 3.33(10 ) ) −6

−6

______________________________________________________________________________________ SOLUTION (5.4)

I y= 2[ 3012×8 + 240 × 342 ] + 8×1276 = 85(104 ) mm 4 3

I z= 2[ 8×1230 + 240 ×192 ] + 7612×8 = 21.3(104 ) mm 4 3

I yz = [0 + 240(19)(−34)] + [0 + 240(−19)(34) + 0] = −31(104 ) mm 4 Using Eq. (5.14):

M o [21.3 + 1.5(−31)]z= M o [−31 + 1.5 × 85] y

(CONT.) ______________________________________________________________________________________ 134 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 5.4 (CONT.) or z = −3.83 y Thus, point A is the farthest from the N.A., as shown. A

N.A.

y Hence, Eq. (5.13) gives

[21.3+1.5( −31)]0.03−[ −31+1.5(85)]( −0.034)

6 = (σ x ) A 80(10 = )

[85×21.3− ( −31)2 ]10−8

from which

= M o 269.1 N ⋅ m

______________________________________________________________________________________ SOLUTION (5.5) Bending moment at the midspan is

Mz = −24(4) + 24(2) = −48 kN ⋅ m

From Fig. 5.4 and Example 5.1:

= z E 0.105 = m zD 0 yE = −0.045 m yD = −0.045 m

I= I= 11.596(10−6 ) m 4 y z I yz = −6.79(10−6 ) m 4 Then, Eq. (5.13) with M = 0 :

( −48000)(11.596)( −0.045) (σ x ) D = 0−[(11.596) = −283.5 MPa 2 − (6.79)2 ]10−6

Similarly,

(σ x ) E =

−48000[ −6.79(0.105) −11.596( −0.045)]

103.8 MPa = [(11.596)2 − (6.79)2 ]10−6

______________________________________________________________________________________ SOLUTION (5.6) Y

P 60 mm 80 mm

z 20 mm

A Y

N . A. for α = 15o N . A. for α = 30o

20 mm

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.6 (CONT.)

= z

80×20(10) + 60×20(50) 80×20 + 60×20

= 27.14 mm

×20 ×60 I y = 8012 + 20 × 80(17.14) 2 + 2012 + 20 × 60( 22.86) 2 3

= 15.1048(105 ) mm 4 ×30 + 2[ 2012 + 20 × 30(25) 2 ]= 8.933(105 ) mm 4 Due to the symmetry I yz = 0.

I z=

( a ) We have

80×203 12

α = 0o

My = 0

M z = 1.5P

Equation (5.13) is thus, 290 (106 ) 1.2

5 P ( 0.04 ) = 1.5I zPy = 81..933 (10− 7 )

P = 3.6 kN

( b ) Now we have α = 15 and o

M z = 1.5P cos 15o = 1.4489 P M y = 1.5P sin 15o = 0.3882 P

Equation (5.14):

0.3882 P (8.933) z = 1.4489 P(15.1048) y

from which

z = 6.311 y

The farthest point from the N.A. is A. Equation (5.13): 6

) 0.3882 P (8.933)( −0.02714) −1.4489 P (15.1048)0.04 (σ x ) A = − 290(10 = 1.2 8.933(15.1048)10−7

Solving, P = 3.36 kN ______________________________________________________________________________________ SOLUTION (5.7) Given α = 30 and o

M z = 1.5P cos 30o = 1.299 P

M y = 1.5P sin 30o = 0.75P The area properties are already found in Solution of Prob. 5.6. Equation (5.14) gives

0.75P(8.933) z = 1.299 P(15.1048) y z = 2.9287 y

As before, the maximum stress occurs at A. Equation (5.13): 6

) 0.75 P (8.933)( −0.02714) −1.299 P (15.1048)0.04 − 290(10 = (σ x ) A = 1.2 8.933(15.1048)10−7

Solving,

P = 3.37 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.8) N.A A z

y We have M y = 0

M z = PL

and

Equation (5.14) becomes

I yz z = I y y

− th 3 z = 23 th 3 y

y = − 3z2

Point A is the farthest from the N.A. Thus, with y A = −h − 2t and z A = − 2t Equation (5.13) yields 3

(σ x ) A = PL[ − (th2 th( −3 t32)()8−th( 23th3)−3()−hth(3−)h2 −t 2 )

(σ x ) A = 3 PL (72th.53t + 2 h ) = σ max

______________________________________________________________________________________ SOLUTION (5.9)

A 80 mm

N.A

I= y

bh3 36

80(903 ) 36

I= z

hb3 48

90(803 ) 48

= 1.62(106 ) mm 4

= 0.96(106 ) mm 4

My = −3(103 ) sin 20o = −1.026 kN ⋅ m

30 mm 90 mm

= M z 3(103 = ) cos 20o 2.819 kN ⋅ m

( a ) Equation (5.15):

96 φ = tan −1 [ 10..62 tan( −20o )] = −12.17 o

( b ) Point A is the farthest from the N.A. Equation (5.16):

−0.03) −0.04) σ A = −1026( − 2819( = 136.5 MPa 1.62(10 ) 0.96(10 ) −6

−6

______________________________________________________________________________________ SOLUTION (5.10) y My

D 60 mm z A

C 100 mm 120 mm

M=10 kN∙m 80 mm

40o

Mz B

1 (120 × 803 − 100 × 603 ) 12 = 3.32 ×106 mm 4 1 (80 ×1203 − 60 ×1003 ) I= y 12 = 6.52 ×106 mm 4 = Iz

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.10 (CONT.)

Mz = −10 cos 40o = −7.66 kN ⋅ m,

My = 10sin 40o = 6. 43 kN ⋅ m

M y z A 7.66 ×103 (−0.04) 6.43 ×103 (0.06) M y = + 3.32 ×10−6 6.52 ×10−6 Iz Iy = −92.29 + 59.17 = −33.12 MPa

− z A+ σA =

σB = −92.29 − 59.17 = −151.5 MPa σ D = 92.29 + 59.17 = 151.5 MPa ______________________________________________________________________________________ SOLUTION (5.11) y

z My

M=4 kN∙m

Iz =

h1 z

A C

② b2

h2 = Iy

①

b1 B

= z

1 (b)(h3 ) 12

1 b (h)(b3 ) + bh( ) 2 12 2 1 = (h)(b3 ) 3

Az (50 × 200)(25) + (100 × 50)(100) ∑ = = 50 mm (50 × 200) + (100 × 50) ∑A

1 1 (50)(200)3 + (100)(50)3 = 34.375(10−6 ) m 4 12 12 1 1 I y = (200)(50)3 + (50)(100)3 = 25(10−6 ) m 4 3 3 Iz =

y A = 25 mm yB = −100 mm zA = −100 mm z B = 0 = = Mz M cos α 4= cos 35o 3.277 kN ⋅ m

= My M = = sin α 4sin 35o 2.294 kN ⋅ m M y Iz

M y zA

M y Iz

M y zB

(a)

− z A+ σA =

(b)

− z B+ σB =

(3.277)(0.025) (2.294)(−0.1) = − + = −11.56 MPa 34.375(10−6 ) 25(10−6 ) (3.277)(−0.1) (2.294)(0) 9.53 MPa = − + = 34.375(10−6 ) 25(10−6 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.12) (a)

σ x = ∂∂yΦ = −c4 xy − 23 c5 x 3 y − c6 xy 3 2

σ y = ∂∂xΦ = c2 x − c3 xy − 23 c5 xy 3 2

τ xy = − ∂∂x∂Φy = c1 + c2 x 2 + c2 y 2 + c5 x 2 y 2 + c4 y 4 2

= 0,

∂ 4Φ ∂y 4

= −6c6 xy ,

and

∂ 4Φ ∂x 4

Thus,

∇ 4 Φ = 6c6 xy + 8c5 xy = 0

∂ 4Φ ∂x 2∂y 2

= −4c5 xy

6c6 + 8c5 = 0 At y = h 2 : σ y = 0 : or

(a)

5h c2 − c23h − c12 =0 At y = − h 2 : σ y = − px Lt :

(b)

( c2 + c3 y − 2 c35 y ) x = − px Lt or

5h c2 + c23h + c12 = − Ltp

(c)

Adding Eqs. (b) and (c),

c2 = − 2pLt

Substituting this into Eq. (b): 3

c3h 2

5h + c12 = − 2pLt On y = ± h2 , τ xy = 0 :

(d)

c6 − c1 − c23 x 2 − c84 h 2 − c45 h 2 x 2 − 64 h4 = 0 2

6h ( c23 + c54h ) x 2 + ( c1 + c48h + c64 )=0 This is of form A ⋅ B + C = 0, where A, B, C are independent. Thus,

c3 2

+ c54h = 0

(f) 2

(e)

6h c1 + c48h + c64 =0 Multiply Eq. (f) by h and subtract it from Eq. (d) to find c5 = − 3Ltp

(g)

Then, Eqs. (g) and (a) give p c3 = 23htL ,

On x = 0 : h

c6 = th43pL

V = 0: h

∫ τ tdy = ∫ (−c − 2

−h 2

c4 y 2 2

− c64y )tdy = 0

h )=0 c1 + c4 ( ) + c6 ( 320 h2 24

(h)

Subtracting Eq. (h) from (g), together with the value of c6 already obtained, we have p c4 = − 53htL

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.12 (CONT.) Finally, substitute c4 and c6 into Eq. (g) to determine

c1 = 80phtL

The stress function is thus, 3

3 3

x Φ = Ltp [ 80h xy − 12 − x4 hy + 10xyh + xhy3 − 5xyh3 ] 3

and the stresses are

σ x = Ltp [ 53h xy + h2 x 3 y − h4 xy 3 ] 3

σ y = Ltp [ − 2x − 23h xy + h2 xy 3 ] 3

2 3

3y 3x y τ xy = Ltp [ 80h − 34xh + 10 − hy ] h + h 2

px σ x = McI = ( px th6 L12)( h 2 ) = Lth

(b)

( c ) The maximum stresses are 3

(σ x ) elast . = 10pth [6 y + 2000 y − 40h 2y ] 3

(σ x ) elem. = p10(10hh3t) = 100( pt ) Thus,

(σ x ) elast . = 99.8( pt ) (σ x ) elast . = 0.998(σ x ) elem.

at y = ± h2

______________________________________________________________________________________ SOLUTION (5.13) 4

( a ) We can show that given Φ satisfies ∇ Φ = 0. From Eqs. (3.13):

σ x = 0.p43 {2[0.78 − tan −1 xy ] − x 2+xyy } 2

(a)

σ y = 0.p43 {2[0.78 − tan −1 xy ] − 2 + x 2+xyy } 2

(b)

τ xy = − 0.p43 x2+y y 2

(c)

To test which stress is maximum we rewrite σ y in the form: 2

σ y = 0.p43 {2[0.78 − tan −1 xy ] − 2 (xx ++yy ) + x 2+xyy } 2

= 0.p43 {2[0.78 − tan −1 xy ] − x 22+xyy 2 − 2x( 2x+− yy2) } Comparing this with Eq. (a), noting 2( x − y )

( x 2 + y 2 ) > 0, we conclude that σ x > σ y .

When y = 0, Eqs. (a) to (c) yield σ x = 3.63 p, σ y = − p 0.43 , τ xy = − p 0.43 Maximum stress occurs at y = 0 :

σ x ,max 3.63(2) = = 7.26 MPa (b)

(σ x )elem=. Thus,

Mc I

1 px 2 ( x ) 2 2 3

= 3= p 6 MPa

x 3

(σ x ) elast . = 1.21(σ x ) elem.

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.14) b σ bottom σ top

= cc12 = 3

y = c1

c1 = 22.5 mm 15 mm c2 90 mm c2 = 67.5 mm Ay 15 mm y = c1 = ∑ Ai i Thus y i ∑ (15)(90)(45) + ( b −15)(15)(7.5) = = 22.5 mm (15)(90) + ( b −15)(15) and

60,750 + ( b −15)(112.5) 1,350 + ( b −15)(15)

= 22.5 mm

Solving, b = 150 mm ______________________________________________________________________________________ SOLUTION (5.15) pL ( th 2 )( h 4 ) 3 pL τ max = VQ Ib = 2 ( th 12 ) t = 4 ht 3

σ max = McI = pL8 thh 212 = 43 pL th 3

t h

L σ max τ max

Thus,

from which

= L Then,

L h

σ max h τ max

8.4(0.15) 0.7

= 1.8 m

p = 43 τ maxL( th ) = 43 700( 0.105.8×0.15) = 3.89 kN m

______________________________________________________________________________________ SOLUTION (5.16) P 4500

R A = P2 − 2250

RB = P2 + 6750

V(N)

4500

0.75P-3375

(N ⋅ m)

P 2

+ 2250 x

6750

We have 3

I = 0.2(0.25) − 0.15(0.2) = 160.417(10−6 ) m 4 12 12 Then,

6 τ = VQ Ib = 0.7(10 )

( P 2) + 2250 160.417×10−6 (0.05)

(0.2 × 0.025 × 0.1125 + 0.1× 0.05 × 0.05)(2)

(CONT.) ______________________________________________________________________________________ 141 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 5.16 (CONT.) Solving, P = 9.32 kN Similarly,

3375 )( 0.125 ) σ = McI = 7(10 6 ) = ( 0.75160P −.417 ×10 −6

or Thus,

P = 16.478 kN

Pall = 9.32 kN

______________________________________________________________________________________ SOLUTION (5.17) y

pL2/2

Equation (5.43) becomes, at a distance x:

bh 2 px 2 2 = 6 σ all At the fixed end ( x = L and h = h0 ):

(1)

bh02 pL2 2 = 6 σ all

(2)

Divide Eq. (1) by Eq. (2),

h 2 h02 = x 2 L2

h = h0

x L

______________________________________________________________________________________ SOLUTION (5.18) We have

Mx =

1 1 1 pLx − px 2 = p ( Lx − x) 2 2 2

Then, at a distance x, Eq. (5.43) becomes

1 p ( Lx − x) bh 2 = σ all 6 At x = L 2 : 1 2 2 bh0 2 p ( L 4) = σ all 6 2

(1)

(2)

Divide Eq. (1) by Eq. (2),

h 2 Lx − x 2 = 2 h02 L 4

or h 2h0 =

x x − ( )2 L L

______________________________________________________________________________________ 142 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (5.19) y

σ all = 150 MPa bh3 1 and Iz = − (b − 2t )(h − 2t )3 12 12 1 1 = (120)(170)3 − (100)(150)3 12 12 6 = 21.005(10 ) mm 4 (a)

Therefore

σ all =

M C

Mc , Iz

t b

I 21.005(10−6 ) 6 150(10 ) 37.1 kN ⋅ m = = σ all c 0.085 1 M 37,100 = = = 0.0252 9 rx EI (70 ×10 )(21.005)(10−6 ) rx = 39.68 m M =

or (b)

______________________________________________________________________________________ SOLUTION (5.20)

1 π ( D 4 − d 4= ) (604 − 404 ) = 510.509(103 ) mm 4 64 64

I =

30 = mm 0.03 m My A (600)(0.03) = σA = = 35.3 MPa 510.509(10−9 ) I

(a)= yA

20 = mm 0.02 m MyB (600)(0.02) = σB = = 23.5 MPa 510.509(10−9 ) I

(b) = yB

(c)

1 M = rx EI

600 70(10 )(510.509 ×10−9 ) = 0.0168 =

M C

B A

rx = 59.52 m

d D

Hence,

r 59.52 − x = − = −205.24 m rz = 0.29 ν

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.21)

= y

A y 2A y + A y ∑ = 2A + A ∑A i

2(150 × 25)(75) + (350 × 25)(137.5) = c2 2(150 × 25) + (350 × 25) z c1 = 108.65 mm So c1 = 108.65 mm and c2 = 41.35 mm

25 25

C 25

150 mm

350 mm

1 = I z 2[ (25)(150)3 + (25 ×150)(33.65) 2 ] + 12 1 (350)(25)3 + (350 × 25)(28.85) 2 = 30.29(106 ) mm 4 12 1 1 M= M = PL = p (2.4) = 1.2 P z 2 2 Therefore

Mc1 1.2 P(0.10865) ; 60(106 ) , P 13.9 MPa = = 30.29(10−6 ) Iz Mc2 1.2 P(0.04135) 60(106 ) , P 36.6 kN = σc = = 30.29(10−6 ) Iz

σt =

Hence Pall = 13.9 kN ______________________________________________________________________________________ SOLUTION (5.22)

M = max

1 2 1 2 = pL (12)(3) = 13.5 kN ⋅ m 8 8

1 3 (80)(120) = 11.52 ×106 mm 4 12 Mc 13.5 ×103 (45 ×10−3 ) = = 52.7 MPa σ= nom I 11.52 ×10−6 σ all 95 = K = = 1.8 σ nom 52.7 D 120 Use Fig. D.3. For= K 1.8, = = 1.33 : d 90 r = 0.14 d

= I

Thus,

= rmin 0.14(90) = 12.6 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.23) p (a) h L pL/2 pL/2 b V, kN

pL/2 x

M kN ⋅ m

-pL/2

Mmax= pL2/8

3 V 3 pL 2 3 pL = = 2 A 2 bh 4 bh 2 Mc pL 8(h 2) 3 pL2 σ= = = max I bh3 12 4 bh 2

τ= max

Thus,

(1) (2)

τ max σ max = h L

(3)

For example, if L = 10h , the above ratio is 1 10 . (b)

From Eq. (3), we have:

σ all 9 0.16(= ) 1.029 m = L h= 1.4 τ all

Equation (1) gives then

p= all

4 bh 4 0.05 × 0.16 6 (1.4 ×10= ) 14.51 kN m τ= all 3 L 3 1.029

______________________________________________________________________________________ SOLUTION (5.24) P

300 mm 60

M = 3P n = E s Et = 20

= It

(520)(300)3 12

= 1170(106 ) mm 4 400

y =150 60

z y

The allowable stress in the transformed section is 120 = 6 MPa < 7 MPa 20

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.24 (CONT.) Thus, the stress in the steel is the controlling stress. Hence,

M max = 3Pmax = σ maxc I t 6

−6

(10 ) = 6(10 )(1170 0.15 )

Pmax = 15.6 kN

______________________________________________________________________________________ SOLUTION (5.25) E

n = Ews = 20 3 3 1 1 I= (64.941 + 37.969t ) ×10−6 m 4 12 [(75 + 40t )(225 )] − 12 [75(100 )] =

= M

= pL2

1 8

2 [40(10)3 (3 = )] 45 kN ⋅ m

Therefore, we write

= σs

nMy I

; 140 ×= 106

My I

6 ; 10 ×10 =

20(45×103 )(0.1125)

= , t 17.5 mm

(64.941+ 37.969 t )×10−6

Similarly

σ= w

(45×103 )(0.1125) (64.941+ 37.969 t )×10−6

,= t 11.6 mm

Stress in steel governs: t = 17.5 mm ______________________________________________________________________________________ SOLUTION (5.26) 180 mm

25 kN/m 10

4m 2

pL 8

M max = = It

1 12

25×10 (4) 8

z 300 mm

= 50 kN ⋅ m 3

) 2 (180)(300)3 + 2[ 1260(10 + (1260)(10)(155) = ] 1010.64(106 ) mm 4 12

180 mm

C y

180x7=1260 mm

σ w,max=

Mc It

σ a ,max =

7 Mc It

50×103 (0.15)

= 7.4 MPa

1010.64(10−6 )

7×50×103 (0.16)

= 55.4 MPa

1010.64(10−6 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.27) Equation (5.54) becomes

20 20 ( kd ) 2 + ( kd )( 300 )(1200) − ( 300 )(500)(1200) = 0

or (kd ) Solving,

+ 80kd − 40(103 ) = 0

kd = 164 mm

Hence,

500 − kd = 336 mm

From Eqs. (e) of Example 5.8:

M c = 12 σ c (bkd )( d − kd3 )

6 0.164 1 = 2 (12 × 10 )(0.3 × 0.164)(0.45 − 3 ) = 131.5 kN ⋅ m and M s = σ s As ( d − kd3 )

= 150(10 6 )(1200 × 10 −6 )(0.445) = 80.1 kN ⋅ m Thus,

M all = 80.1 MPa

______________________________________________________________________________________ SOLUTION (5.28) 5 C

d A 80/8

E c1=kd D B

The stresses in concrete and the equivalent of the steel (Fig. 514b) have the values shown in the preceding figure. From the similarity of ECD and EAB we find c1 d

= (80 58)+5 = 155

c1 = 515d = 155

Equation (d) of Example 5.8:

1 2

σ c ( c1b) = σ s As

1 2

(5)( 500 3 )300 = 80 As

As = 1563 mm 2

Then, we have from Eq. (e) of Example 5.8,

M = σ s As ( d − c31 )

= 80(10 6 )(1563 × 10 −6 )(0.5 − 09.5 ) = 55.6 kN ⋅ m ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.29) y D

Vy z

From of Example 5.9:

2 3 th + 2bth 2 3 2 = (4)(90)3 + 2(100 × 4)(90) 2 3 = 8.424 ×106 mm 4 = Iz

h C

h b (a)

Equation (e) of Example 5.9 with t1= 2

t2= t :

= e

3 b 3 (100) = = 38.46 mm 2 h + 3b 2 90 + 3 ×100

(b) = τD

bh 5 ×103 (0.1)(0.09) V= = 5.34 MPa y I 8.424 ×10−6

(c)

The first moment of the shaded section about the z-axis:

1 Qz = bt (h) + ht (h 2) = ht (2b + h) 2

Shear stress formula is thus:

τ max = =

Vy Qz Vy (ht 2)(2b + h) = 2 2 I zt th (6b + h)t 3 3Vy (2b + h)

(P5.29)

4th(6b + h)

Substitute the given data

3(5 ×103 )(2 × 0.1 + 0.09) 4,350 = τ max = 4(0.004 × 0.09)(6 × 0.1 + 0.09) 0.994 ×10−3 = 4.38 MPa ______________________________________________________________________________________ SOLUTION (5.30) The shearing stress at a distance s is given by V yQz

2V y πRt

sin α

τ = I b = I ∫ R cos θ (tRdθ ) =

This shows that τ = 0 at the free ends and τ max at the neutral axis, same as for a

rectangular section. The shearing stress produces the following twisting moment about O: (CONT.) ______________________________________________________________________________________ 148 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 5.30 (CONT.)

T = ∫ τRdA = ∫

π 2V sin α y

πRt

R(Rtdα ) =

4 RV y

By applying the principle of moment at O: V y e = M .

Thus,

e= π

______________________________________________________________________________________ SOLUTION (5.31) Shearing stress in the web is neglected. Moment of the forces about S: e1 e2

V1e1 = V2 e2 ;

= VV12

(a)

Let M 1 and M 2 be bending moments on flanges 1 and 2, respectively. Curvature-moment are related by 1 r1

M1 = EI , 1

M2 = EI 2

(b)

dM 1 I1

(c)

1 r2

By assuming r1 = r2 , we have M1 EI1

M2 = EI ; 2

V1 V2

= II12

e1 e2

= II12

= dMI 2 2

Introducing dM dx = −V , Eq. (c) gives Equation (a) now becomes Since Thus, where

[ ( h −e1e1 ) ] = II12

e1 + e2 = h; e1 = ( I1I+2hI 2 ) 3

I 1 = b121 t1

I 2 = b122 t2

______________________________________________________________________________________ SOLUTION (5.32) A 1

C y

Fig. (a) Location of centroid C (Fig. a):

z =

15.5×75( −37.5) + 0 + 9.5×125(62.5) 15.5×75 + 9.5×250 + 9.5×125 15.5×75( −250) + 0 + 9.5×250( −125) 15.5×75 + 9.5×250 + 9.5×125

= 6.481 mm = −124.339 mm

Moment of inertia:

I y = 121 (15.5)(75) 3 + (15.5 × 75)( 43.98) 2 + 121 ( 250)(9.5) 3 + 250(9.5)(6.48) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.32 (CONT.)

+ 121 (9.5)(125) 3 + 9.5(125)(56.01) 2

= 8.18(106 ) mm 4 I z = 121 (75)(15.5) 3 + (75 × 15.5)(125.6) 2 + 121 (9.5)( 250) 3 + (9.5 × 250)(0.65) 2 + 121 (125)(9.5) 3 + (125 × 9.5)(124.3) 2

= 49.10(106 ) mm 4 I yz = (15.5 × 75)( −125)( −43.98) + (9.5 × 250)( −0.65)( −6.48) + (9.5 × 125)(124.3)(56) = 14.70(106 ) mm 4 .7 ) o θ p = 12 tan −1 [− 8.218(14−49 .1 ] = 17.85

Then,

I y' = 106 [ 8.18+2 49.1 + 8.18−2 49.1 cos 35.7o − 14.7 sin 35.7o ] = 3.45(106 ) mm 4

Similarly, we compute

53.85(106 ) mm 4 I= I= z' 1

I= I= 3.45(106 ) mm 4 y' 2 From geometry of section (Fig. b):

HB 144.57 mm BC 39.05 mm = = Thus,

Vy '

τ xz = I t = [ st (0.1445 − 2s sin 17.85o ] z'

Shear force due to V y ' (Fig. b): s

We write

V ( 0.0155 )

F1 = ∫ τ xz tds = 54y ' .1×10−6 ∫ 0

0.075

[0.1445s − s2 sin 17.85o ]ds = 0.1108V y ' 2

V y ' ⋅ ez ' = 0.25F1 = 0.25(0.1108V y ' )

or= ez '

0.0277 = m 27.7 mm

Shear force due to Vz ' . Now assume that the

direction of F1 shown in the figure is reversed. Then,

F1 = VIzy''t ∫

0.075

We write

[0.03905s − s2 cos 17.85o ]ds 2

= 0.1928Vz '

s 2

144.57

Vy’ Vz’ ey’

Vz ' ⋅ e y ' = 0.25(0.1928Vz ' )

C B

17.85

z’ z

y y’ ez’

Figure (b)

= ey ' 0.0482 = m 48.2 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.33) 1 2

x 3

( p0 Lx ) x

p0 Lx

M A = p03L

RA =

p0 L 2

We have EIv = M = 16 p0 x

Integrating

− 12 p0 Lx + 13 p0 L2 2

EIv ' = 24p0L x 4 − p40 L x 2 + po3L x + c1

v ' (0) = 0;

c1 = 0. 2

EIv ' = 24p0L x 4 − p40 L x 2 + po3L x

(a)

Integrating

0 EIv = 120pEIL x 5 − p120 L x 3 + po6L x 2 + c2 v (0) = 0; c2 = 0.

(a)

0 ( x 5 − 10 L2 x 3 + 20 L3 x 2 ) v = 120pEIL

11 p L4

( b ) Let x=L in this equation: v B = 1200EI p L3

( c ) Let x=L in Eq. ( a ): θ B = 80EI

______________________________________________________________________________________ SOLUTION (5.34)

EIv ' ' ' = px + c1 px + c1 x + c2

EIv ' ' ' ' = p

EIv ' ' =

1 2

EIv ' = 16 px 3 + 12 c1 x 2 + c2 x + c3 EIv = 241 px 4 + 16 c1 x 3 + 12 c2 x 2 + c3 x + c4 Boundary conditions:

EIv (0) = 0; EIv ' ' (0) = 0;

c4 = 0 c2 = 0

EIv ( L) = 0;

pL3 24 3

+ c16L + c3 = 0 3

(a)

EIv ' ( L) = − EI 96pLEI = pL6 + c12L + c3 3

c3 = − 1796pL − c12L

(b)

Substituting Eq. (b) into (a), we obtain reaction at right end:

c1 = − 1332pL = R

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.35) P

MA RA

b=L-c

MB B

y Segment AD:

EIv1 ' = M A x − 12 R A x 2 + c1

EIv1 ' ' = M A − R A x

EIv1 = 12 M A x 2 − 16 R A x 3 + c1 x + c2

(a)

Segment BD:

EIv 2 ' ' = M A − R A x + P( x − c ) EIv 2 ' = M A x − 12 R A x 2 + 12 P ( x − c ) 2 + c3 EIv 2 = 12 M A x 2 − 16 R A x 3 + 16 P( x − c ) 3 + c3 x + c4

Boundary conditions:

v1 (0) = 0; v1 ' (0) = 0;

and

(b)

c2 = 0 c1 = 0

v1 ( c ) = v 2 ( c ); v1 ' ( c ) = v2 ' ( c );

c3 c + c 4 = 0 c3 = 0, c4 = 0

v2 ' ( L) = 0 = M A L − 12 R A L2 + 12 Pb 2

(c)

v2 ( L) = 0 = 12 M A L2 − 16 R A L3 + 16 Pb 3

(d)

Solving Eqs.(c) and (d), 2

R A = PbL3 (3c + b) = P ( LL−3 c ) ( 2c + L) 3

Substituting of this into Eq. (c) gives

M A = Pcb = Pc ( LL2−c ) L2 2

Thus, introducing the values of M A , R A , and c1 = c2 = c3 = c4 = 0 into Eqs. (a) and (b)

we obtain the deflections. For 0 ≤ x ≤ c : 2

c) x v1 = P ( L6 −EIL (3cL − 2cx − Lx ) 3

For c ≤ x ≤ L : c) x v2 = P ( L6 −EIL (3cL − 2cx − Lx ) + 6PEI ( x − c ) 3 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.36)

M y

L-x

The reactions are statically indeterminate. We have

EIv ' ' = M = RB x − RB L + M 0

EIv ' = 12 RB x 2 − RB Lx + M 0 x + c1

v ' (0) = 0;

c1 = 0.

EIv = 16 RB x 3 − 12 RB Lx 2 + 12 M 0 x 2 + c2

v (0) = 0;

c2 = 0

v ( L) = 0;

RB = 32ML0

The preceding equation gives 2

( x−L) v = M 0 x4 EIL

______________________________________________________________________________________ SOLUTION (5.37) p0 L 4

Owing to symmetry: RA = RB =

For 0 ≤ x ≤ L 2 :

M A = −M B

Thus

M = R A x − M A − p30Lx = p04Lx − M A − p30Lx EIv" = − M A + p04Lx − p30Lx 2

EIv' = − M A x + p0 8Lx − p120 xL + c1

Boundary conditions:

θ = v ' ( 0) = 0 :

c1 = 0

= v '( L2 ) 0= : MA Consequently

2 2

5 p0 L2 96

0L x EIv = − 5 p192 + p024Lx − p600 xL + c2

Boundary condition:

v ( 0) = 0:

Therefore

= v and

p0 x 2 960 LEI

c2 = 0

(−25 L3 + 40 L2 x − 16 x 3 )

v= v= ( L2 ) max

7 p0 L4 3840 EI

( for 0 ≤ x ≤ L 2 )

↓

______________________________________________________________________________________ 153 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (5.38) r

M1 M2

1 2

( a ) Static equilibrium gives

P1 = P2 = P

M 1 + M 2 = Ph

Equation (5.9):

M 1 = E1rI1

M 2 = E2rI 2

Interface strains must be the same:

α1∆T +  due to temp . increase

Fig. (a) Bent beam

due to axial force

= α 2 ∆T − EP22h − hr2

E1h 

r 

due to bending ( from Eq . 5.9 )

This yields an expression for the interface curvature: 1 r

= 12h ((14α2+−nα+11) ∆nT)

where n = E1 E 2 .

( b ) At the interface

σ 1 = Ph + E2 rh = (σ yp )1 1

σ 2 = − Ph − E2 rh = −(σ yp ) 2 2

( c ) Summing these equations.

(σ yp )1 − (σ yp ) 2 = − h ( E21 −r E2 )

It follows that

(σ yp )1 −(σ yp ) 2 E1 − E2

= h2 24h ((14α 2+−nα+11) ∆nT)

from which

+ n +1 n ∆T = 14 6 (α 2 −α1 )

(σ yp )1 −(σ yp ) 2 E1 − E2

______________________________________________________________________________________ SOLUTION (5.39) Case B h

b t

h y

Q = Qoutside − Qinside = b2 ( h 2 − y12 ) − b−22 t [( h − t ) 2 − y12 ] = bht − 12 bt 2 + th 2 + t 3 − ty12

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.39 (CONT.) Q dA α = IA ∫ ( width ) 2

bt  h −t ( bht +th 2 − bt −2 ht 2 +t 3 −ty12 )2  + th 2 − 2 ht 2 + t 3 −ty12 ) 2 h ( bht − 2 2 tdy bdy 2 [ 2 ] +  ∫0 1 1  ∫h−t b2 b2   A α = Aweb = 1 + 2 ( hb−t ) 2

A I2

or where

A = 2bh − (b − 2t )( 2h − 2t ) = 2bt + 4th − 4t 2 Aweb = 2( 2h − 2t )t = 4ht − 4t 2 I = 23 bh 3 − 23 (b − 2t )( h − t ) 3 b Case B

Q = ( h − t − y1 )t[ y1 + 12 ( h − t − y1 )] = th2 − ht 2 + t2 + bht − bt2 − 2t y12 2

bh bt ty1 t 2  h −t ( bht + ht 2 − bh − bt − by1 + t )2  − − + ) h ( bht + ht 2 + 2 2 2 2 2 2 2 2 A  ] α = I 2 2[ ∫ tdy bdy + 2 2  1 1 ∫h−t b t  0  b A or α = Aweb = 1 + h −t 2

where

A = 2bt + ( 2h − 2t )t = 2bt + 2th − 2t 2 Aweb = ( 2h − 2t )t = 2ht − 2t 2 I = 121 t ( 2h − 2t ) 3 + 2[ 121 bt 3 + bt ( h − 2t ) 2 ]

Note: For thin-walled sections flange and web thicknesses are small with compared with to unity and their products are neglected. Case C O θ/2 r

r sin θ2

y = r cos θ2 A*

We write A = r2 (θ − sin θ ) =area of segment 2

dA = r2 (1 − cos θ )dθ = r 2 (1 − cos 2 θ2 )dθ 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.39 (CONT.)

Q = ∫ * ydA = ∫ * ( r cos θ2 ) r 2 (1 − cos 2 θ2 )dθ A

= r ∫ (cos 2 − cos 3 θ2 )dθ = r 3 [2 sin θ2 − 23 sin θ2 (cos 2 θ2 + 2)] 3

0 3

= r [sin θ2 − sin θ2 cos 2 θ2 ] 2 3

Q α = IA ∫ ( width dA ) 2

A I2

∫

2π [

2 πr 3 θ θ θ (sin −sin cos2 )] 3 2 2 2

( 2 r sin ) 2 2

A = πr 2

Since,

r 2 (1 − cos 2 θ2 )dθ

I 2 = π16r

2 8

We have, after simplification: 2π

α = π16r ∫ ( r9 )(1 − cos 2 θ2 ) 3 dθ = 109 6

Case D

θ/2

φ/2 y

Q = Q2 − Q1 = 23 r22 [sin θ2 − sin θ2 cos 2 θ2 ] − 23 r12` [sin φ2 − sin φ2 cos 2 φ2 ]

α = IA [ ∫ 2

Letting

2π

Q22

( 2 r2 sin ) 2

2π

Q12

( 2 r1 sin ) 2 2

r 2 (1 − cos 2 θ2 )dθ − ∫

θ 2 2

A = π ( r22 − r12 ),

after integration, we obtain

r12 (1 − cos 2 φ2 )dφ ]

I 2 = π16 ( r24 − r14 ) 2

α =2

______________________________________________________________________________________ SOLUTION (5.40) Since stress is symmetrical, Eq. (3.40) reduces to

( ∂∂r 2 + 1r ∂∂r )( ∂∂rΦ2 + 1r ∂∂Φr ) = ∂∂rΦ4 + 2r ∂∂rΦ3 − r12 ∂∂rΦ2 + r13 ∂∂Φr = 0 2

The given Φ satisfies this equation. Equations (3.32) lead to

σ r = 1r ∂∂Φr = rA + B(1 + 2 ⋅ ln r ) + 2C 2

σ θ = ∂∂rΦ = ( − rA ) + B(3 + 2 ⋅ ln r ) + 2C 2

The constants A, B, and C are determined from the boundary conditions (b) and (d) of Sec. 5.13. In so doing, we arrive at the solution given by Eq. (5.67). ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.41) 46 mm 6 mm

z z 30 mm

6 mm

y A2

= z

( A1 z1 + A2 z2 ) ( A1 + A2 )

= 16.14 mm

We have

= r 136.14 = mm ri 120 = mm ro 166 mm A 420 R= = 126 =134.7045 mm, e = r − R =1.4355 mm 166 dA ∑ ∫ r ∫ 30rdr + ∫ 6rdr 120

126

We have M = −(30 + 120 + 16.14) P = −166.14 P Outer Edge. Applying Eq. (5.73),

166.14 P (134.7045 −166) P −80 mmN 2 = 420 + 420(1.4355)(166)

or= P

1.614 = kN Pall

Inner Edge. Using Eq. (5.73), Solving,

80 mmN= 2

P (134.7045 −126) + 166.14 420(1.4355)(126)

P 420

P = 3.74 kN

______________________________________________________________________________________ SOLUTION (5.42) P M B A

Figure (a) ( a ) Equation (5.66) yields

N = (1 − 00..212 ) 2 − 4 00..212 ln 2 00..21 = 0.082 2

M =PR =70(103 ) × 0.15 =10.5 kN ⋅ m Thus, from Eq. (5.67):

, 500 ) (σ θ ) A = 0.054( 0(10 [(1 − 00..212 )(1 + ln 1) − (1 + 1) ln 00..21 ] .2 ) 2 ( 0.082 ) 2

= −163 MPa 2 2 , 500 ) (σ θ ) B = 0.054( 0(10 [(1 − 00..212 )(1 + ln 2) − (1 + 00..212 ) ln 2] .2 ) 2 ( 0.082 ) = 103 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.42 (CONT.) Referring to Fig. (a),

( b ) We have

σ max =(σ θ ) A − PA =−163 − 705 =−177 MPa σ min = (σ θ ) B − PA = 103 − 14= 89 MPa

ln 2 144.2695 = ri 100 mm, = ro 200 mm, = R 100 =

= A 5000 = mm 2 , e 5.7305 Thus,

− PA [1 + ( eri i ) ] = −176.2 MPa (σ θ ) A = r R −r (

)

r R −r (σ θ ) B = 88.1 MPa − PA [1 + ero o ] =

______________________________________________________________________________________ SOLUTION (5.43)

A = 20b + 2(60 ×10) = 20b + 1200 We have rA = 60 mm and rB = 140 mm . Applying Eq. (5.70): (− M )( R − rA ) (− M )( R − rB ) σB = −σ A = = − AerA AerB from which

rB ( R − rA ) = −rA ( R − rB ) , 140( R − 60) = −60( R − 140) or R = 84 mm A Then R= dA ∑∫ r 140 2(10) dr  8 14    80 bdr or 84 b ln + 20 ln  = A 84  ∫ +∫ =  80 6 8 r    60 r = 24.1653b + 940.1545 Hence 20b + 1200 = 24.1653b + 940.1545 or b = 62.4 mm ______________________________________________________________________________________ SOLUTION (5.44)

1 We have c = d = 10 mm r = b + c = 25 mm 2 1 R = [r + r 2 − c 2 ] (by Table 5.2) 2 1 = [25 + 252 − 102 ] = 23.9564 mm 2 e = r − R =1.0436 mm P 2 = A π= c 2 π (10) = 314.16 mm 2

r O

P M=-P(a+ r ) B

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.44 (CONT.) (a)

From Eq.(5.74):

(a + r )( R − rA ) P M ( R − rA ) P = [1 + ] A AerA A erA 800 (50)(23.9564 − 15) [1 + ]= = 75.4 MPa 314.16 (1.0436)(15)

σA = −

(b)

Using Eq.(5.74):

(a + r )( R − rB ) P M ( R − rB ) P = [1 + A AerB A erB 800 50(23.9564 − 35) [1 + ]= = −35.9 MPa 314.16 1.0436(35)

σB = −

______________________________________________________________________________________ SOLUTION (5.45) Locate centroid : 150 mm

50 mm

A1r1 + A2 r2 A1 + A2 C A1 O 75 mm A2 (5625)(150) + (3750)(200) = = 170 mm 5625 + 3750 100 mm 1 A =(50 + 75)(150) = 9375 mm 2 2 r r=

1 2 h (b1 + b2 ) 2 R= r (b1ro − b2 ri ) ln o − h(b1 − b2 ) ri (0.5)(150) 2 (75 + 50) = 158.9163 mm 25 [(75)(250) − (50)(100)]ln − (150)(75 − 50) 10 P e = r − R =11.0837 mm (a)

Use Eq.(5.75a),

(σ θ ) A = −

r ( R − rA ) P M ( R − rA ) P ] − = − [1 + A AerA A erA

50(103 )  (170)(158.9163 − 100)  = − 1+ 9375  (11.0837)(100)  = −53.5 MPa

B M=P r

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.45 (CONT.) (b)

From Eq.(5.75b):

r ( R − rB ) P − [1 + (σ θ ) B = ] A erB

50(103 )  (170)(158.9163 − 250)  = − 1+ 24.5 MPa = 9375  11.0837(250) 

______________________________________________________________________________________ SOLUTION (5.46)

1 A = bh 2

The section width w varies linearly with r. Thus (1) w= c0 + c1r

Since

= w b= (at r ri ) = w 0= (at r ro )

(2)

Substituting Eq.(1) into Eq.(2);

bro b c1 = − c0 = h h

ro w c0 + c1r dA dr dr = ∫A r ∫= ri r r Inserting c1 and c0 into this, after integrating and rearranging,

b ri rο

Then

we have

r r dA = ∫ r b( ho ln roi − 1)

Therefore

= R

1 h A 2 = dA ro ro ∫ r h ln ri − 1

______________________________________________________________________________________ r SOLUTION (5.47) r

A = π c2

Through use of the polar coordinates we write:

w= 2c sin α r= r − c cos α dr = c sin α dα = dA wdr = 2c 2 sin 2 α dα

dA π 2c 2 sin 2 α ∫ r =∫0 r − c cos α dα

c ccosα

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.47 (CONT.) 2 2 2 2 2 2 2 π c (1 − cos α ) π r − c cos α − ( r − c ) = 2= 2 ∫0 r − c cos α ∫0 r − c cos β π π dα = 2 ∫ (r + c cos α )dα − 2(r 2 − c 2 ) ∫ 0 0 r − c cos α

This gives

= 2r α 0 + 2c sin α 0 − 2(r 2 − c 2 )

r −c

r 2 − c 2 tan

tan −1

r +c

2 0

∫ r = 2π (r − r − c ) 2

Hence, it can be shown that

A 1 = (r + r 2 + c 2 ) dA 2 ∫r

______________________________________________________________________________________ SOLUTION (5.48)

= A

1 (b1 + b2 )h 2

The section width w varies linearly with r as We have

w b2

w= c0 + c1r

(1)

= w b= (at r ri ) 1

rο

(2)

= w b= (at r ro ) 2

Introduce Eq.(1) into Eq.(2), then solve for c0 and c1

c0 =

rob1 − rb i 2 h

Then, we write

b −b c1 = − 1 2 h

c0 + c1r r dr= c0 ln o + c1 (ro − ri ) ri r ri

∫ r = ∫ r dr= ∫ ri

This gives, substituting Eqs.(3):

(3)

r dA r b − rb = ∫ r o 1 h i 2 ln roi − (b1 − b2 )

Hence

1 2 h (b1 + b2 ) A 2 = R = dA r ∫ r (rob1 − rbi 2 ) ln roi − (b1 − b2 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (5.49) P z M A2 A B A1

z A2

P We have ( A1 z1 + A2 z2 ) ( A1 + A2 )

= 12.5 mm

z =

= r 52.5 = mm A 0.0008 = m2 ri 40 = mm ro 80 mm

∑∫

dA r

800

∫

50 dr 40 r

+∫

2(5) dr r 50

= 50.4502 mm, e = r − R = 2.0498 mm

M = (0.06 + 0.04 + 0.0125) P = 337.5 N ⋅ m Inner Edge. Using Eq. (5.74), −3

337.5(50.4502 − 40)10 3000 (σ θ ) A = − 0.0008 − 0.0008(2.0498)(40)10 −6

= −57.5 MPa Outer Edge. Applying Eq. (5.74), −3

337.5(50.4502 −80)10 3000 (σ θ ) B = − 0.0008 − 0.0008(2.0498)(80)10 −6

= 72.3 MPa ______________________________________________________________________________________ SOLUTION (5.50) Using Eq. (P5.51), at section A-B of Fig. P5.50:

M= −0.182 Pr = −0.182(0.75b) P = −0.136 Pb

( a ) Equation (5.66) yields,

N = (1 − 14 ) 2 − 4( 14 )(ln 2) 2 = 0.082

Point A. Applying Eq. (5.67),

(σ θ ) A = tb4 2MN [(1 − 14 )(1 + ln 1) − 2 ⋅ ln 2] + btP Pb ) 3 = 4tb( −20(.0136 [ 4 − 1.3863] + btP .082 )

= 5.22 P bt Point B. Using Eq. (5.67),

Pb ) 3 (σ θ ) B = 4tb( −20(.0136 [ 4 (1 + ln 2) − 45 ln 2] + btP .082 )

= − 1.68P bt (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.50 (CONT.) b ) ln 2 Thus, R (= 0.75 = b, ri 0.5b, and ro = b . = 0.7213b 2 Points A and B. Apply Eq. (5.74) with P = P 2 and M = −0.136 Pb : P P (σ θ ) A = 5.2 P bt bt + 4.2 bt = P P −1.64 P bt (σ θ ) B = bt − 2.64 bt =

( b ) We= have r

(σ θ ) A,B = ( 2PA ) ± ( McI )

(c)

Here

c = b4

I = tb96

Thus,

0.136 Pb ( b 4) P P P = (σ θ ) A = 4.26 P bt bt + bt + 3.26 bt = tb3 96

and

(σ θ ) B = btP − 3.26 btP = − 2.26 P bt

Comparing the result, we observe the following differences: At the point A: Elasticity vs. hyperbolic 0.4 % Elasticity vs. linear 18 % At the point B: Elasticity vs. hyperbolic 2.4 % Elasticity vs. Linear 34.5 % ______________________________________________________________________________________ SOLUTION (5.51) From the condition of symmetry, the distribution of stress in any quadrant is known to be the same as the other.

P 2

r θ

cos θ

Ma P/2

At any angle θ , the bending moment referring to this figure is expressed as follows Mθ = M a − 12 Pr (1 − cos θ )

(a)

The problem is statically indeterminate and value of M a is found first. Note that shear component of the load will be omitted in our solution. Applying Castigliano’s theorem, we have: ∂U ∂M a

= 4∫

π 2

1 EI

M θ ∂∂MMθa ds

= 0

4 EI

∫

= 0

4 EI

[ π2 M a − Pr2 ( π2 − 1)]

[ M a − Pr2 (1 − cos θ )]rdθ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.51 (CONT.) solving,

M a= Pr ( 12 − π1 = ) 0.182 Pr

Then, Eq. (a) becomes

M θ= 0.182 Pr − 12 Pr (1 − cos θ )

(b)

Therefore stress at any point of a section, using Eq. (5.74), is expressed in the form

σθ = − P2 cosAθ + MAθ ( Rer− r )

Here, the moment M θ is given by Eq. (b). ______________________________________________________________________________________ SOLUTION (5.52) ( a ) Using Eq. (P5.51) at θ = π 4 (Fig. P5.51): π = M θ 0.182(0.15) P − 12 P(0.15)(1 − cos = 0.00533P N ⋅ m 4)

We have

= A 0.05(0.1) = 0.005 m 2

= R

and

h 100 = = 144.2695 ro ln ri ln 12

e = r − R =150 − 144.2695 = 5.7305 mm At inner fiber, ri = −0.1 m : −3

−3

− ( P 2)cos(π 4) P (144.2695×10 ) − (100×10 ) (σ θ )π 4 = − 0.00533 −153.06 P Pa 0.005 0.005 [ (5.7305×10−3 )(100×10−3 ) ] =

(b)

Nθ Mθ Vθ O

θ (Mθ)θ=0 P/2

r At θ = 0

: M θ = 0.182 Pr

Nθ = − P 2

At any angle:

Mθ = 0.182 Pr − 0.5 Pr (1 − cos θ ) = −0.318 Pr + 0.5 Pr cos θ ∂M θ ∂ ( P 2)

= −0.636r + r cos θ

N θ = − P2 cos θ , Vθ = − P2 sin θ ,

∂Nθ ∂( P 2) ∂Vθ ∂( P 2)

= − cos θ

= − sin θ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 5.52 (CONT.) Based on the symmetry of the ring, we have

= δ 2∫

π 2 N

∂Nθ θ AE ∂ ( P 2)

Vθ ∂Vθ + MEIθ ∂∂( MP θ2) + AG ∂ ( P 2) ]rdθ

Substituting the values of M θ , N θ , and Vθ , this expression becomes

= δ P 2∫

π 2

2 α P P [ AE cos 2 θ + 2PrEI (−0.636 + cos θ ) 2 + AG 2 sin θ ]rdθ 2

π 2

=2[ 2PrAE θ2 + 14 sin 2θ 0 + 2PrEI 0.904θ − 1.27 sin θ + sin42θ 0 3

π 2

Pr θ 1 + 2αAG 2 − 4 sin 2θ 0 ]

from which

α Pr π δ P = 2[ 2PrAE ( π4 ) + 2PrEI (0.15) + AG 2 ( 2 )] 3

For the given rectangular cross section:

= r 0.15 = m A 0.005 m 2 α = 65 = 1.2 G = 25 E = 0.4 E = I

= (0.05)(0.1)3 4.17(10−6 ) m 4

1 12

The deflection is therefore 3

( 0.15 )π P ( 0.15 ) ( 0.15 ) P ( 0.15 )π 1.2 δ P = 2[ 0P.005 ] 8 E ( 8 ) + 2 E ( 4.17×10 ) + 0.4 E ( 0.005 ) −6

This results in

δ P = 215.65P E m End of Chapter 5

CHAPTER 6 SOLUTION (6.1)

τ all = c

or 6

π 2

or 100(10 )[1.5(10 Solving,

−6

(c 4 − b 4 )

100(106 ) =

2(3 ×103 )(0.035) π [(0.035) 4 − b 4 ]

) − b4 ] = 66.845

= b 0.0302 = m 30.2 mm

______________________________________________________________________________________ SOLUTION (6.2) D t = 25

23 Di = 25 D

d=40 mm

= τ max or

T (d 2) T ( D 2) = 4 π d 32 π ( D 4 − Di4 ) 32

23 4 ) ] 25 (40)3 = D 3 (0.2836) D = 60.9 mm

= d 3 D 3 [1 − (

______________________________________________________________________________________ SOLUTION (6.3)

Di = D2 d

τ= max

16T = πd3

T ( D 2) 16T = 1 π ( D 4 − Di4 ) π D 3 (1 − ) 32 16

3 15 = d 3 D= ( ), D 1.0217 d 16

Thus

= d

D 60 = = 58.7 mm 1.0217 1.0217

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.4) Apply the method of sections between the change of load points:

= TCD 0.5 kN ⋅ m = TAB 2 kN ⋅ m

= TBC 1.5 kN ⋅ m

Therefore, τ max = 2T π c : 3

= τ AB

2(2 ×103 ) = 81.5 MPa π (0.025)3

= τ BC

2(1.5 ×103 ) 119.4 MPa = π (0.02)3

2(0.5 ×103 ) 94.3 MPa = τ CD = π (0.015)3

______________________________________________________________________________________ SOLUTION (6.5) We have, applying the method of sections:

= TCD 0.5 kN ⋅ m = TAB 2 kN ⋅ m

Hence,

τ max = gives,

= τ AB

= TBC 1.5 kN ⋅ m

2Tc π (c 4 − b 4 )

2(2 ×103 )(0.025) = 82.4 MPa π [(0.025) 4 − (0.008) 4 ]

2(1.5 ×103 )(0.02) = 122.5 MPa τ BC = π [(0.02) 4 − (0.008) 4 ] = τ CD

2(0.5 ×103 )(0.015) = 102.6 MPa π [(0.015) 4 − (0.008) 4 ]

______________________________________________________________________________________ SOLUTION (6.6)

TBC = 3 kN ⋅ m

TAB = 2 kN ⋅ m

Then,

τ max = 2T π c3 yields

= τ BC

2(1×103 ) = 79.58 MPa π (0.02)3

2(2 ×103 ) = 81.49 MPa τ AB = π (0.025)3 Therefore,

(τ max = ) B K= τ AB 1.6(81.49) = 130.4 MPa

______________________________________________________________________________________ 167 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (6.7) We have Then

= TBC 4 kN ⋅ m

= TAB 4 kN ⋅ m

φC = ∑ (TL GJ ) yields 4(103 )

0.35 0.7 1 0.35 0.7 ]= ( ) − − 4 3 π 3.5π 6.25 150 (28 × 109 ) (0.05) (0.1) 4 2 2 −3 o = 4.67 ×10 rad = 0.27

φC =

[

______________________________________________________________________________________ SOLUTION (6.8)

θ = 50 + 90 = 140o π 4

J =[(0.05) − (0.045) 4 ] = 3.376 ×10−6 m 4 2 o From Eq. (6.5a) at θ = 140 : σ x ' = τ sin 280o = −0.985τ Tc T (0.05) 6 120 ×10 0.985= 0.985 = J 3.376 ×10−6 or = T 8.23 kN ⋅ m Similarly, Eq. (6.5b):

cos 280o 0.174τ = τ x ' y ' τ=

gives 6 50 ×10= 0.174

Thus,

T (0.05) , 3.376 ×10−6

= T 19.4 kN ⋅ m

= Tall 8.23 kN ⋅ m

______________________________________________________________________________________ SOLUTION (6.9) From solution of Prob. 6.8:

= J 3.376 ×10−6 m 4

From Equations (6.5) at θ = 35 + 90 = 125 , o

σ x ' = τ sin 250o = −0.94τ 120 × (106 )= 0.94

or = T and

8.62 kN ⋅ m

Tc T (0.05) = 0.94 J 3.376 ×10−6

τ x ' y ' = τ cos 250o = −0.342τ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.9 (CONT.)

T (0.05) , 3.376 ×10−6

6 50 ×10= 0.342

Thus,

= T 9.87 kN ⋅ m

= Tall 9.87 kN ⋅ m

______________________________________________________________________________________ SOLUTION (6.10) 3 kN ⋅ m

2 kN ⋅ m

30 mm

Apply the method of sections:

= TAB 2 kN ⋅ m

Then,

1 kN ⋅ m

= TBC 1 kN ⋅ m

2 ×103 J π d13 T = = = 16 τ all 50 × 106 c = d1 0.05884 = m 58.84 mm

Also,

φ = ∑ (TL GJ ) :

103

1(1) 2(2) − 4 ] 4 π (0.04) d1 (39 ×109 ) 32 4 76576.32 = 390625 − 4 d1

0.02 =

Solving

[

d1 =59.74 ×10−3 m =59.74 mm

We therefore use

d1 = 60 mm

______________________________________________________________________________________ SOLUTION (6.11) 40 mm

2 kN m

1.5 kN ⋅ m

A 3m

0.5 kN ⋅ m

Apply the method of sections:

= TBC 0.5 kN ⋅ m

= TAB 1.5 kN ⋅ m

(CONT.) ______________________________________________________________________________________ 169 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 6.11 (CONT.) We have

J π d13 T 1.5 ×103 = = = 16 τ all 50 ×106 c

Also,

d1 =53.46 ×10−3 m =53.5 mm

φ = ∑ (TL GJ ) : 103

0.5(1) 1.5(3) − 4 ] 4 (0.04) d1 9 π (40 ×10 ) 32 4.5 78539.82 = 195,312.5 − 4 d1

or 0.02 =

Solving Use

[

d1 =78.79 ×10−3 m =78.8 mm

d1 = 78.8 mm

______________________________________________________________________________________ SOLUTION (6.12) (a)

D 60 = = 1.2 d 50

r 1 = = 0.02 d 50

Figure D.4: K = 2.1 .

2T 2(2 ×103 ) = = 81.5 MPa π c3 π (0.025)3 = τ max 2.1(81.5) = 171.2 MPa

τ= nom

r 5 D = = 0.1, = 1.2; = K 1.33 d 50 d = τ max 1.33(81.5) = 108.4 MPa (b)

(Fig. D.4)

______________________________________________________________________________________ SOLUTION (6.13) F C

125 mm

B 190 mm

= TCD F (0.125) = 600 N ⋅ m or

F = 4.8 kN

= TAB 4.8(0.19) = 912 N ⋅ m

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.13 (CONT.) (a)

φ AB=

= φCD Thus ( b ) τ= AB

TL GJ

912(1.5)

= 0.0279 rad

80(109 ) (0.025)4 2 600(1.8)

= 0.0537 rad π

(80×109 ) (0.02)4 2

φD = φCD + 1.52φ AB = 0.0963 rad = 5.52o 2TAB

2(912)

= 37.16 MPa

π c3

π (0.025)3

______________________________________________________________________________________ SOLUTION (6.14) (a) and

= TAC

9549(8) 1500

= 50.93 N ⋅ m

9549 kW n

50.93 T c3 = , τ all = 210×106 2.2

Similarly

c= 6.98 mm

9549(25) 1500

= 159.2 N ⋅ m

= TCB 3

π = 2 c

159.2 210×106 2.2

,= c 10.2 mm

( b ) We have φ = TL GJ , with J = πc Thus = φ AC

π 2

φBC =

π 2

Hence

∴ DAC = 13.96 mm

∴ = DCB 20.4 mm

50.93(2)

= 0.333 = rad 19.08o

(0.00698)4 (82×109 ) 159.2(5)

0.571 rad 32.72o = = 4 9

(0.0102) (82×10 )

φ AB = φBC − φ AC = 13.64o

______________________________________________________________________________________ SOLUTION (6.15) ( a ) Use Eq.(4.8a): σ

yp = n

Thus,

32 M 16T σ x2 + 4τ xy2 with σ x = + π4dP , τ xy = πd πd 3

σ yp

π d3

[(8M + Pd ) 2 + (8T ) 2 ]

1 2

(P6.15a)

Inserting the data, we find 250(106 ) n

4(103 )

[(8 × 4 + 40 × 0.08) 2 + (8 × 8) 2 ] 2 π (0.08) 3

n = 1.38

( b ) Similarly Eq.(4.9a) becomes: σ yp n

So,

= σ x2 + 3τ xy2 = π 4d 3 [(8M + Pd ) 2 + 48T 2 ] 2 6

250(10 ) n

4(10 )

π (0.08)3

[(8 × 4 + 40 × 0.08) 2 + 48(8) 2 ]

(P4.15b)

1 2

or n = 1.53 ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.16)

= T and π

9549(12) 3000

38.2 140×106 2.5

9549 kW n

3 c=

φall Also = L

τ all

= 38.2 N ⋅ m

c 7.57 mm = 38.2(2)

π = ; 1.8( 180 )

= , c 9.92 mm

T GJ

80(109 )π c 4

A 20-mm dia. shaft should be used. ______________________________________________________________________________________ SOLUTION (6.17) Equation (6.4a) gives

= T (a)

9549kW 9549(10 0.7457) = = 21.34 N ⋅ m 6000 n

Equation (6.1) with c = d 2 and J = π d

16T d ; = πd3

= τ all

32 :

16T

πτ all

Substitute the data

d = (b)

16(21.34) m 11.6 mm = 0.0116 = π (70 ×106 )

By Eq.(6.3):

TL 32TL = GJ π Gd 4 32(21.34)(0.6) rad 10.58o = = 0.1847 = π (39 ×109 )(0.0116) 4

φ =

______________________________________________________________________________________ SOLUTION (6.18)

r 10 = = 0.2 d 50

D 100 = = 2.0; = K 1.25 (from Fig. D.4) d 50

d= 2 25 mm : 16T 80(106 ) = 1.25[ ] π (0.05)3

For smaller part of the shaft,= c

K( τ max = τ= all Solving

16T ); πd3

= T 1.571 kN ⋅ m

______________________________________________________________________________________ 172 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (6.19) b

b a

( a ) For circular bar:

θ c = G2πTb

τ c = π2bT = Gθ c b

For elliptical bar: 2

θ e = Tπ(aa b+bG ) ,

T = πaa2 +bbG2 θ e 3 3

3 3

τ e = π2abT = 20aθ +bab G e 2

We have τe τc

= ( a 22+θbeba2 )θGbG c

Setting θ e = θ c : τe τc

= ( a22 +a b2 )

Since a > b, ( a

+ b 2 ) < 2a 2 , and τ e τ c > 1, or

τe >τc 4

Tc = θ cπ2b G ;

(b)

Gθ c b = τ c

τ c πb G τ cπb Tc = Gb 2 = 2 4

3 3

Te = θ eaπ2a+bb2G ,

τ e = 2aθ ba+b G e 2

Rearranging,

θ e = τ 2( aba+Gb ) e

Thus,

Te = τ eπ2ab

We obtain

Tc Te

= ((ττ cππabb2 22)) e

Setting τ e = τ c :

Tc Te = b a , or

Te > Tc

______________________________________________________________________________________ SOLUTION (6.20) We have π ( c (a)

− b 2 ) = πa 2 ; 4

a 2 = c 2 − b2 .

Th = Jτc a = πτ a ( 2c c−b ) Ts = Jτaa = πτ 2a a

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.20 (CONT.) T

Hence, Th = c −b3

c 4 −b4 c ( c 2 −b2 )

For c = 1.4a : Th Ts

(b)

8416 = 12..3168 = 2.16 4

Th = JGLφa = ( GLφa )[ π ( c 2−b ) ]

Ts = ( GLφa )( πa2 ) 4

Hence, Th = c a−2b

= ( cc2 −−bb2 )2 = cc2 +−bb2

For c = 1.4b : Th Ts

96 = 02..96 = 3.08

______________________________________________________________________________________ SOLUTION (6.21) Use Eqs. (f) of Example 6.2:

TA = 1+( aJTb bJ a ) =

T 0.4 (15 ) 4 (π 32 )

= (1+0.T6328) = 0.6124T

0.2 ( 20 ) 4 (π 32 )

TB = T − TA = 0.3876T Based on shear in segment AC:

T) τ a = 16πdT = 16π( 0( .06124 = 150(10 6 ) .02 ) A 3 a

or= T

384.7 N ⋅ m

Based on shear in segment CB: or

16TB πd b3

T) = 16π((00.3876 = 150(106 ) .015 )3

= T 256.5 N= ⋅ m Tall

______________________________________________________________________________________ SOLUTION (6.22) Use Eqs. (f) of Example 6.2:

= TA

T 0.8(15)4 (π 32) 1+ 0.5(25)4 (π 32)

= 0.8282T

T 1+ 0.2074

TB =T − TA =0.1718T Based on shear in segment AC: 16TA = π d3 a

16(0.8282T )

70(106 ) = π (0.025)3

= T 259.3 N= ⋅ m Tall

Based on shear in segment CB: 16TB = π d3

16(0.1718T )

= 70(106 ) π (0.015)3

= T 270 N ⋅ m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.23) ∂ 2Φ ∂x 2

= k [2( a 2 b − a 2 ) + 2 y 2 (b 2 + 1) − 12bx 2 ]

∂ 2Φ ∂y 2

= k [2( a 2 b − a 2 ) + 2 x 2 (b 2 + 1) − 12by 2 ]

Substituting these in Eq. (6.9):

k = 2 a 2 ( b−1)+( b2G−θ6b+1)( x 2 + y 2 )

In order k be a constant: b

Gθ 2 a 2 ( b −1)

− 6b + 1 = 0 . Thus,

______________________________________________________________________________________ SOLUTION (6.24) The condition of compatibility requires that 2

∇ 2φ = ∂∂xφ2 + ∂∂yφ2 = H = constant However for given stress function: 2 ∇= φ

2k a 2b 2

[a 2 + b 2 − ( x 2 + y 2 )] ≠ constant

So, Equation P6.24 cannot be used for a Prantl stress function. ______________________________________________________________________________________ SOLUTION (6.25)

u = −θz ( y − b), ε x = ε y = ε z = 0,

v = θz ( x − z ), w = w( x, y ) ∂v ∂u γ xy = ∂x + ∂y = 0

γ xz = ∂∂wx − θ ( y − b), Thus

γ yz = ∂∂wy + θ ( x − a )

τ xz = G[ ∂∂wx − θ ( y − b)] = ∂∂Φy τ yz = G[ ∂∂wy − θ ( x − a )] = − ∂∂Φx

Substituting these in Eqs. (6.6), (6.7), and (6.11), we obtain ∂τ xz ∂y

and

∂τ

− ∂xyz = −2Gθ ,

∂ 2Φ ∂x 2

+ ∂∂yΦ2 = −2Gθ 2

T = ∫∫ [( x − a )τ yz − ( y − b)τ xz ]dxdy = [ − ∫∫ ( x − a ) ∂∂Φx − ∫∫ ( y − b) ∂∂Φy ]dxdy = 2 ∫∫ Φ dxdy

We observe that characteristic equations remain unchanged. ______________________________________________________________________________________ SOLUTION (6.26)

T = ∫∫ ( xτ yz − yτ xz )dxdy = − ∫∫ ( x ∂∂Φx − y ∂∂Φy )dxdy

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.26 (CONT.) Here

− ∫ dy ∫ x ∂∂Φx dx = − ∫ xdΦ dy − [ − ∫∫ Φ dxdy ] x1

= −c ∫ ( x 2 − x1 )dy + ∫∫ Φ dxdy = −c ∫∫ dxdy + ∫∫ Φ dxdy Note that ( x 2 − x1 )dy is area and equals

Thus,

∫∫ dxdy . Similarly,

− ∫ dx ∫ y ∂∂Φy dy = −c ∫∫ dxdy + ∫∫ Φ dxdy T = 2 ∫∫ (Φ − c )dxdy

______________________________________________________________________________________ SOLUTION (6.27)

Tθ = T cos 2 θ Hence, θ dz θ = ∫ TGJ = GJ1 ∫ T cos 2 θadθ

This gives, at sections A and B: Ta θ A = GJ ∫

π 2

( 12 + cos22θ )dθ = 2Ta r 4G

2 Ta θ B = GJ ∫ cos θdθ = rTaG 4

______________________________________________________________________________________ SOLUTION (6.28) b=a c 2a

From Table 6.2: α = 0.246, β = 0.229 T τ r = αabT = 0.492 a τ c = TcJ = π2cT 2

Thus,= T 0.492 a3

= ; c 0.679a

2T π c3

Similarly,

θ r = 0.229 (T2 a ) a G = 0.458T a G 3

T θ c = JG = π2cTG 4

Then,

T 0.458 a 4G

= π2cT4G ; = c 0.735 = a call

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.29) ∂ 2Φ ∂y 2

= k [( x + h3 ) + ( x + 3 y − 23 h ) + ( x + h3 ) + ( x − 3 y − 23 h ) + 2( x − 23 h )]

∂ 2Φ ∂y 2

= k [ −3( x + h3 ) − 3( x + h3 )]

Substituting these in Eq. (6.9),

− 4kh = −2Gθ ;

Thus,

k = G2 θh

Φ = −Gθ [ 12 ( x 2 + y 2 ) − 21h ( x 3 − 3xy 2 ) − 272 h 2 ]

Along the x axis τ xz = 0, due the symmetry. Equation (6.8) is therefore,

for y = 0 : When

τ yz = − ∂∂Φx = 32Ghθ ( 23hx − x 2 ) x = 0: τ yz = 0 x = 23h : τ yz = 0 h x=−3: τ yz = τ max = G2θh

Next, substitute the preceding value of Φ into Eq. (6.11) to obtain

T = 2 ∫∫ Φ dxdy = −4Gθ ∫

∫

− 3 y + 23 h

[ 12 ( x 2 + y 2 ) − 21h ( x 3 − 3xy 2 ) − 272 h 2 ]dxdy

Gθh = 15 3

This gives θ = 15 4 3T h G

The shear stress is thus

τ max = ( 2 h20T3 ) = 152 h3T 3

______________________________________________________________________________________ SOLUTION (6.30) For a circular bar:

T = GθJ = C cθ

where

C c = πr2G

(a)

For an elliptical bar (from Example 6.4): 2

H = − 2T π( aa 3b+3b ) = −2Gθ

T = aπ2a+bb2 Gθ = C eθ 3 3

3 3

C e = aπ2a+bb2 G

(b)

For an equilateral bar (from Prob. 6.29): Gθh T = 15 = θC t 3 4

Ct = 15h G3 Bars have equal areas:

Ac = πr 2 ,

(c)

Ae = πab,

At = h 3

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.30 (CONT.) Setting Ac = Ae = At :

r 4 = a 2 b 2 = 3hπ 2 4

Then, Eqs. (a), (b), and (c) give Ce 2 a 3b3 Cc r 4 ( a 2 + b 2 ) Ct Cc

and

= a22 +abb2

= 15h G3 πr24G = 2π15 3

______________________________________________________________________________________ SOLUTION (6.31) do

di t

For the seamless tube, from θ 1 = T GJ

θ1 = πd (1−32d T d )G 4 o

4 i

4 o

For the split tube, referring to Eq. (6.17):

θ 2 = 3GT

1 d + di do −di 3 )( ) 2 2

π( o

We have θ1 θ2

= 32((dd2o+−ddi2)) o

(a)

For very thin tubes d o + d i θ1 θ2

≈ 2d o2 , and Eq. (a) becomes

= 43 ( dto )

______________________________________________________________________________________ SOLUTION (6.32) Apply Eqs. (6.17) and (6.19):

τ max=

θ =

3T bt 2

3T bt 3G

3(80)

= 76.8 MPa

0.125(0.005)2 3(80)

= 0.192 rad m

0.125(0.005)3 (80×109 )

______________________________________________________________________________________ SOLUTION (6.33) Referring to Table 6.2, we have

= φ θ= L

Thus

φ τ

β ab3G

τ max =

α ab 2

= bGL αβ

(1)

For a b = 24 16 = 1.5 : α = 0.231 and β = 0.196 Introducing given data into Eq. (1): 1.5(π 180 )

τ max

Solving,

4 ( 0.231) = ( 0.024 )(0.80 ×109 )( 0.196 )

τ max = 106.6 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.34) From Eq. (1) of solution of Prob. 6.33: L α b = τ max G φ β

(a)

By Table 6.1: α = 0.208 and β = 0.141. Substituting the numerical values into Eq. (a), we obtain 120(106 )

= 10.1 mm

= b

0.204 2 79(109 ) 25(π 180) 0.141

______________________________________________________________________________________ SOLUTION (6.35) Equation (6.20) yields

bt ∑=

= Je

1 3

(100)(10)3 + 13 (125)(4)3

= 3.6(104 ) mm 4 Maximum shear stress occurs on the lower leg: Tt1 Je

τ max=

500(0.01)

= 138.9 MPa

3.6(10−8 )

Angle of twist per unit length is

= θ

500 3.6(10−8 )(200×109 )

T J eG

= 69.44(10−3 ) rad m = 3.98o per meter ______________________________________________________________________________________ SOLUTION (6.36) Refer to Table 6.2. ( a )= T Also

a23τ max 20

(0.045)3 (50×106 ) 20

= 228 N ⋅ m

) T = a246Gθ.2all = ( 0.045) (80×4610.2 )(1.5π 180= 186 N ⋅= m Tall

(b)

.5 φ = 46.2GT [ aL + aL ] = 4680.2×(10186 ) [ ( 0.206.5) + ( 0.1045 ] ) max

1 4 1

2 4 2

= 1.0742(10−7 )[192,901.235 + 365, 797.897] rad 3.44o = 0.06 = ______________________________________________________________________________________ SOLUTION (6.37) Referring to Fig. P6.37, an expression for t is written as

t = t0 (1 − by )

Substitute this into the given stress function to obtain, 2

We have

Φ = Gθ [ t40 (1 − by ) 2 − x 2 ] T = 2 ∫∫ Φ dxdy = 4 ∫

t0 2

= 4Gθ ∫

t0 2

b (1− 2t x ) t 2 0 0 4 0

∫

[ (1 − by ) 2 − x 2 ]dxdy

t02 4

{ [b(1 − 2t0x ) − b(1 − 2tox ) 2 + b3 (1 − 2t0x ) 3 ] − bx 3 (1 − 2t0x )}dx

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 6.37 (CONT.) Let u =

(1− 2 x ) t0

x = (1−2u ) to

dx = − t02du

du = − 2tdx0 ,

Then, the preceding expression for the torque becomes 1

T = 4G t08b ∫ (u 2 − 23 u 3 )du 0

Integrating,

T = Gθt03 12

______________________________________________________________________________________ SOLUTION (6.38) ( a ) From Table 6.2: T = ( 2πr tG )θ = Cθ 3

C = 2πr 3tG = 2π (0.02375) 3 (0.0025)G = 2.1(10 −7 )G T T τ= = = 112,863T max 2π r 2 t 2π (0.02375)2 (0.0025)

( b ) Equation (6.16): 3

C = bt3G = ( 0.02375)(30.0025) G = 1.24(10 −10 )G 3

Equation (6.18):

τ max = bt3T = ( 0.023753)(T0.0025) = 20,210,526T 2

( c ) From Table 6.2: for a = b, t = t1 :

T = (0.02375) 3 tGθ = 1.34(10 −5 )0.0025Gθ = 3.35(10 −8 )Gθ = Cθ τ max = 2 Ta 2t = 2 ( 0.02375T)2 ( 0.0025) = 354,571T ______________________________________________________________________________________ SOLUTION (6.39) Referring to Table 6.2: 420(106 )(2) 3

τ A = 20a T ; 3

Solving, = T Also

= θ

20T = (0.05) 3

1.75 kN ⋅ m 46.2(1.75×103 )

= 0.1617 rad m

46.2T a 4G

(6.25×10−6 )(80×109 )

______________________________________________________________________________________ SOLUTION (6.40) For a thin-walled tube, letting

ro ≈ ri ≈ ravg . = r 4

J = π ( ro2− ri ) = π2 ( ro2 + ri 2 )( ro + ri )( ro − ri ) = 2πrt

Equation (6.2):

ρ τ = TJρ = 2Tπρr t = 2TArt 3

Since r = ρ , we obtain Eq. (6.22): τ = T 2 At . ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.41) For a regular hexagon, we can write

A = 3 23 a 3

ds = 6a

Equation (6.22), substituting the value of A, results in the shear stress

τ = 2TAt = T9 a 3t 2

Angle of twist per unit length, using Eq. (6.23): 6a ) θ = τ2(GA = 23ATaGt 2

2T θ = 9Ga t 3

______________________________________________________________________________________ SOLUTION (6.42) 0.25

0.25 t3

t2 A1 t1

0.25

t2 Given:

t1= 0.012 m,

t2= t3= 0.006 m,

t4= t5= 0.01 m,

G= 28 GPa

T= 56.5 kN ⋅ m, A1 = A2 = A= 0.0625 m 2 s= s= s= s= s= 0.25 m. 1 2 3 4 5 We write or

T = 2 A1h1 + 2 A2 h2 = 2 A1t1τ 1 + 2 A2 t3τ 3 2τ 1 + τ 3 = 75.3(10 6 )

(1)

The shear flow yields

Also,

τ 1t1 = τ 2 t 2 τ 2 t 2 = τ 3t 3 + τ 5 t5 τ 3t 3 = τ 4 t 4

(2) (3) (4)

τ 1 s1 + 2τ 2 s2 + τ 5 s5 = 2GθA1 − τ 5 s5 + 2τ 3 s3 + τ 4 s4 = 2GθA2

(5) (6)

(CONT.) ______________________________________________________________________________________ 181 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 6.42 (CONT.) Simultaneous solution of Eqs. (1) through (6), after substitution of the given numerical values yields:

= = τ 1 19.1 MPa τ 2 38.2 MPa = τ 3 37.2 = τ 4 22.3 MPa MPa and

τ 5 = 0.62 MPa

θ = 6.86(10−3 ) rad m

______________________________________________________________________________________ SOLUTION (6.43) a

t=3.5 mm

a=50 mm We have

= A Then,

(0.05)(0.05sin 60o ) 2

= 1.0825(10−3 ) m 2 .

= h

T 2A

τ=

h t

θ=

2 GA

= 18, 475.751 N m

40 2(0.0010825)

18,475.751 0.0035

∫ ds=

= 5.274 MPa 5.274(106 ) 2(28×109 )(0.0010825)

[0.05 + 0.05 + 0.05]

= 13.05(10−3 ) rad m ______________________________________________________________________________________ SOLUTION (6.44) t

c 1

2 a

For circular tube:

Area enclosed by c, Am1 = πc

Area of the section, A1 = 2πct Polar moment of inertia, J 1 = 2πc t 3

For square tube:

Enclosed area by a, Am 2 = a

Area of the section, A2 = 4at

(CONT.) ______________________________________________________________________________________ 182 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 6.44 (CONT.) Polar moment of inertia (from Table 6.2 with t = t1 and a = b ): 2 2

J 2 = (2attt1+abtb1 ) = a 3t Then, A1 = A2 gives: a = πc 2 and hence, from Eq. (6.22) we obtain τ1 τ2

= AAmm12 = πac 2 = 0.785 2

Similarly, from φ = T GJ : φ1 φ2

= JJ12 = 2aπc3 = 0.617 3

______________________________________________________________________________________ SOLUTION (6.45) t1

t2 A2

t3 t2

t1 We have 1

s1 = [0.252 + 0.03752 ] 2 = 0.2528 m 1

s2 = [0.52 + 0.052 ] 2 = 0.5025 m = s3 0.15 = m, s4 0.05 = m, s5 0.075 m and

τ 1t1 = τ 3t3 + τ 2 t 2 τ 5t5 = τ 1t1 , τ 2 t2 = τ 4 t4 T = 2 A1t1τ 1 + 2 A2 t 2τ 2

(1) (2,3) (4)

Using Eq. (6.23),

τ 5 s5 + 2τ 1 s1 + τ 3 s3 = 2GθA1 2τ 2 s2 + τ 4 s4 − τ 3 s3 = 2GθA2

(5) (6)

t2= t4= t5= 0.0005 m, t3= 0.00075 m, G= 28 GPa, and = T 4 kN ⋅ m.

Given: t1=

The cell areas are calculated as

= A1 0.028125 = m2 A2 0.05 m 2 Substituting the numerical values and solving Eqs. (1) through (6):

τ= τ= 50.6 MPa, 1 5 τ= τ= 50.88 MPa, 2 4

τ= 0.593 MPa 3 θ= 0.0194 rad m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.46) The area and mean length are

A= (3a )2a + π 2a = a2 (12 + π ) S = 2(3a ) + 2a + π a = a (8 + π ) 2

(a) (b)

Equation (6.22) is thus,

= τ

T 2 tA

T ta 2 (12 +π )

The angle of twist per unit length by Eq.(6.23) is determined as

= θ =

= ∫ dst

h 2 AG

4[ a2 (12 +π )]2 Gt

T 2A 2 AGt

[a (8 + π )]

= 0.0486 GaT 3t

The total angle of twist of the tube is then

θ ( L= θ 0.0486 GaTL t ) L= 3

______________________________________________________________________________________ SOLUTION (6.47)

(10) 4 981.748 mm 4 =

(= a) J

φ =

TL GJ

;= 75o 1.309 rad =

T (1.5) 79(109 )(981.748×10−12 )

= T 67.68 N ⋅ m

(b)= τ

16T

16(67.68)

π d3

= 344.7 MPa

π (0.01)3

______________________________________________________________________________________ SOLUTION (6.48) 3

(a) = d

16 PR

π τ all

16(2.4×103 )(0.18)

π (340×106 2)

= d 23.5 mm

( b ) Equation (6.33):

L =

π d 4Gδ

= 32 PR 2

π (0.0235)4 (79×109 )(0.045)

= 1.369 m

32(2400)(0.18)2

______________________________________________________________________________________ SOLUTION (6.49) Angle= of twist, φ

= TL JG ; T φ JG L , where o = φ 15 = 0.262 rad 4 4 = J π= d 32 π (10) = 32 981.75= mm 4 0.98175(10−9 ) m 4

Thus,

= T

0.262(0.98175×10−9 )(79×109 ) 1.1

From Eq.(6.1):

= τ 16T= πd3

= 18.47 N ⋅ m

16(18.47)

= 94.07 MPa

π (0.01)3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (6.50) ( a ) We have δ s = δ c and Eq. (6.33) becomes Ps Rs3 Gs

Solving,

= PGc Rcc ;

Ps ( 0.062 )3 9

79 (10 )

c ( 0.05 ) = P41 (109 )

Ps = 1.011Pc

Assume that copper controls

τ c = 16πPd R ; c c 3

0.05 ) 300(10 6 ) = 16π P(c0(.01 )3

from which Then,

Pc = 1178.1 N Ps = 1191.1 N

Check the assumption:

= τs

16(1191.1)(0.062)

= 376.11 MPa π (0.01)3

Since τ s < 500 MPa, the assumption is correct.

Thus, the total force is

Pt = Pc + Ps = 2369.2 N

( b ) For the condition specified, Ps ks

= kPcc

from which ks kc

= PPcs 1.011 End of Chapter 6

CHAPTER 7 SOLUTION (7.1) Refer to Fig. 7.2 and Eq. (7.10): ∂4w ∂x 4

= h14 ( w9 − 4 w1 + 6w0 + w11 )

∂4w ∂y 4

= h14 ( w10 − 4 w2 + 6w0 − 4 w4 + w12 )

∂4w ∂x 2∂y 2

= h14 [ w5 − w6 + w7 + w8 − 2( w1 − w2 − w3 − w4 + 2 w0 )] 4

Substituting these into ∇ w :

h 4 ∇ 4 w = 20w0 − 8( w1 + w2 + w3 + w4 ) − 2( w5 + w6 + w7 + w8 )

+ w9 − w10 − w11 − w12

______________________________________________________________________________________ SOLUTION (7.2) y

x h=a/2

4a Only a quarter of the section need be considered. Using the symmetry the nodal points are labeled as shown in the preceding figure, and Φ = 0 at the boundary. The finite difference equations for the nodes 1 through 8 are, respectively:

− 4Φ 1 + 2Φ 2 + 2Φ 3 = −2Gθh 2

(1)

Φ 1 − 4Φ 2 + 2Φ 4 = −2Gθh Φ 1 − 4Φ 3 + 2Φ 4 + Φ 5 = −2Gθh 2

(2) (3)

Φ 2 + Φ 3 − 4Φ 4 + Φ 6 = −2Gθh 2

(4)

Φ 3 − 4Φ 5 + 2Φ 6 + Φ 7 = −2Gθh

(5)

Φ 4 + Φ 5 − 4Φ 6 + Φ 8 = −2Gθh

(6)

Φ 5 − 4Φ 7 + 2Φ 8 = −2Gθh 2

(7)

Φ 6 + Φ 7 − 4Φ 8 = −2Gθh 2

(8)

Solving,

Φ 1 = 3.587Gθh 2 Φ 3 = 3.467Gθh 2

Φ 2 = 2.708Gθh 2 Φ 4 = 2.622Gθh 2

Φ 5 = 3.037Gθh 2

Φ 6 = 2.313Gθh 2

Φ 7 = 2.055Gθh 2

Φ 8 = 1.592Gθh 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.2 (CONT.) Tables of differences are in P7.2a and P7.2b. Table P7.2a (when y=b, Φ = Φ 1 ; y=h,

y 0

Φ Gθh 3.587

h 2h

2.708 0

Φ = Φ 2 , etc.) ∆ Gθh 2 ∆2 Gθh 2 − 0.380 − 1.828 − 2.708

Table P7.2b (when x=0, Φ = Φ 1 ; x=h,

Φ Gθh 3.587 3.467 3.037 2.055 0

x 0 h 2h 3h 4h

Φ = Φ 3 , etc.)

∆ Gθh 2 − 0.120 − 0.430 − 0.982 − 2.055

∆2 Gθh 2 − 0.310 − 0.552 − 1.074

∆3 Gθh 2 − 0.242 − 0.522

∆4 Gθh 2 − 0.28

We have ∂Φ ∂x

= ∆Φh 0 + ∆2 hΦ20 [2 x − h ] + ∆6 hΦ30 [3x − 6 xh + 2h 2 ] 4

+ ∆24Φh 40 [4 x 3 − 18 x 2 h + 22 xh 2 − 6h 3 ] 2

( ∂∂Φx ) x =4 h = ∆Φh 0 + ∆2 hΦ20 (7h ) + ∆6 hΦ30 ( 26h 2 ) + ∆24Φh 40 (50h 3 ) Using Table P7.2b, we obtain ( ∂∂Φx ) x =4 h = −1.418Gθa. That is

(τ ) x = ±2 a = −1.418Gθa

Similarly, using Table P7.2a:

( ∂∂Φy ) y =2 h = ∆Φh 0 + ∆2 hΦ20 ( 2 y − h ) = −1.811Gθa

(τ ) y = ±2 h = −1.811Gθa

Hence, the error using this method is [(1.86 − 1.811) 1.86]100 = 2.63 % ______________________________________________________________________________________ SOLUTION (7.3) Equation (7.19) is applied at points b, c, d, respectively:

2Φ c + 1.178Φ g − 4.857Φ b = −2Gθh 2

Φ b + 1.178Φ f + Φ d − 4.857Φ c = −2Gθh 2 1.178Φ c + 1.178Φ e − 5.714Φ d = −2Gθh 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.3 (CONT.) Similarly, Eq. (7.16) is applied at points e, f, and g, respectively:

2Φ d + Φ f − 4Φ e = −2Gθh 2

2Φ c + Φ g + Φ e − 4Φ f = −2Gθh 2 2Φ b + 2Φ f − 4Φ g = −2Gθh 2 In matrix form these equations are written as:

0 0 0 2 − 4 Φ b   2    0 2 0 1 1 Φ c  −4   0 2 1 0  Φ d  −4  0    1.178 − 5.714 1.178 0 0  Φ e   0  1 1 0 1.178 0  Φ f  − 4.857    2 0 0 0 1.178 Φ g  − 4.857 = −2Gθh 2 {1, 1, 1, 1, 1, 1} Solution is

Φ b = 1.612Gθh 2

Φ c = 1.487Gθh 2

Φ d = 0.975Gθh 2

Φ e = 1.547Gθh 2

Φ f = 2.236Gθh 2

Φ g = 2.424Gθh 2

The finite differenceso are then

∆ = Φ e − Φ B = 1.547Gθh 2 o

∆2 = Φ f − 2Φ e + Φ B = −0.858Gθh 2 o

∆3 = Φ g − 3Φ f + 3Φ e − Φ B = 0.357Gθh 2 o

∆4 = Φ f − 4Φ g + 6Φ f − 4Φ B + Φ B = −0.232Gθh 2 o

∆5 = Φ e − 5Φ f + 10Φ g − 10Φ f + 5Φ e − Φ B = −0.018Gθh 2 o

∆6 = Φ B − 6Φ e + 15Φ f − 20Φ g + 15Φ f − 6Φ e + Φ B = 0.036Gθh 2 Hence,

τ B = ( ∂∂Φx ) B = h1 ( ∆ − ∆2 + ∆3 − ∆4 + ∆5 − ∆6 )Gθh 2 0.357 0.232 0.018 0.036 = (1.547 + 0.858 2 + 3 + 4 − 5 − 6 )Gθh 2

= τ B 2.143 = Gθ h 0.0107Gθ

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.4) Applying Eq. (7.16) at points b, c, e, f, and g, we obtain

Φ g + 2Φ c − 4Φ b = −2Gθh 2

Φ b + Φ d + Φ f − 4Φ c = −2Gθh 2

2Φ d + Φ f − 4Φ e = −2Gθh 2

Φ g + Φ e + 2Φ c − 4Φ f = −2Gθh 2

2Φ f + 2Φ b − 4Φ g = −2Gθh 2 At point d, we apply Eq. (7.19):

1.308Φ c + Φ e − 5.77Φ d = −2Gθh 2

Solving these equations, we have

Φ b = 2.238Gθh 2

Φ c = 2.000Gθh 2

Φ d = 1.096Gθh 2

Φ e = 1.715Gθh 2

Φ f = 2.667Gθh 2

Φ g = 2.953Gθh 2

The finite differences are the computed as shown in Table P7.4a. Thus,

τ A = ( ∂∂Φx ) A = h1 ( ∆ − ∆2 + ∆3 − ∆4 )Gθh 2 0.093 0.0186 = ( 2.238 + 1.523 2 + 3 + 4 )Gθh = 3.035Gθh = 0.0129Gθ 2

Table P7.4a

y 0 h 2h 3h 4h

Φ Gθh 2 0 2.238 2.953 2.238 0

∆ Gθh 2 2.238 0.715 − 0.715 − 2.238

∆2 Gθh 2 − 1.523 − 1.430 − 1.523

∆3 Gθh 2 0.093 − 0.003

∆4 Gθh 2 − 0.019

Alternatively, we construct the table as shown in Fig. P7.4b: Table P7.4b

y 0

Φ Gθh 2 2.953

∆ Gθh 2 − 0.715

h 2h

2.238 0

− 2.238

∆2 Gθh 2 − 1.523

Then we obtain, 2

τ A = ( ∂∂Φy ) y =2 h = ∆Φh + 23 ∆ hΦ = [ −0.715 − 23 (1.523)]Gθh = −3.0Gθ h 0

This result is approximately the same as calculated before. Clearly now computation involved are reduced considerably. ______________________________________________________________________________________ 189 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (7.5) y

1.5a A

b=a 1

h=a/4 x

Applying Eq. (7.16) at nodes in to 6, we obtain

0 1 0 0  Φ1  1 − 4 1   1   1 −4 1 0 1 0 Φ2     2 −4 0 0 1  Φ 3   0 2 1    = −2Gθh    0 0 −4 1 0  Φ 4  1  2 1  0 2 0 1 − 4 1  Φ 5      0 2 0 2 − 4  Φ 6  1  0

Solving,

Φ 1 = 1.551Gθh 2

Φ 2 = 2.206Gθh 2

Φ 3 = 2.387Gθh 2

Φ 4 = 1.997Gθh 2

Φ 5 = 2.887Gθh 2

Φ 6 = 3.137Gθh 2

Then, the differences are as shown in Table P7.5. Table P7.5

x 0 h 2h 3h 4h

Φ Gθh 2 0 2.387 3.137 2.387 0

∆ Gθh 2 2.387 0.75 − 0.75 − 2.387

∆2 Gθh 2 − 1.637 − 1.50 − 1.637

∆3 Gθh 2 0.137 − 0.137

∆4 Gθh 2 − 0.274

At point A, we have 2

( ∂∂Φx ) x =0 = h1 ( ∆Φ 0 − ∆ 2Φ 0 + ∆ 3Φ 0 − ∆ 4Φ 0 )Gθh 2

Substituting the first row of Table P7.5: or

0.137 0.274 ( ∂∂Φx ) A = ( 2.387 + 1.637 2 + 3 + 4 )Gθh

( ∂∂Φx ) A = 0.83Gθa = τ max

This differs 2.12 % from the exact solution 0.848Gθa (Table 6.2) ______________________________________________________________________________________ 190 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (7.6) L/2

P v-1

O y

L/2 v2

Boundary conditions yield,

v (0) = 0; v0 = 0

v ' (0) = 0; v1 = v −1

and

Apply Eq. (7.22) at nodes 0, 1, and 2, respectively: 2

v −1 − 2v0 + v0 = h 2(EIPL ) ;

v1 = 4hEI PL

PL 3) v 2 − 2v1 + v0 = h (22EI ;

v 2 = 65hEI PL

3) v3 − 2v2 + v1 = h (2PL EI

(a)

Substituting v1 and v 2 into Eq. (a) we obtain

v3 = 327 PL EI

______________________________________________________________________________________ SOLUTION (7.7) L/2

Mo A

3EI

O y

L/2

B 4

RA M 3Mo/4

Let C = − EI0 h

h=L/4

Mo/2

Mo/4

v0 = v 4 = 0

Applying Eq. (7.22) at 1, 2, 3: o

Solving,

v2 − 2v1 + v0 = C4 v1 = − 83 C

v3 − 2v2 + v1 = C4

v 2 = − 12 C

(a)

θ C = θ 3 = 21h ( v3 − v1 ) = 0

(b)

vC = v2 = 361 MEI0 L

v4 − 2v3 + v2 = C4

v3 = − 83 C

Results are the same as the exact solutions. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.8) 4

Let C = ( p EI )h . We have h = a 4 = L 8 and v0 = v8 = 0. p -1

EI 1

O y

2EI

C L=2a

8 5

-7

Apply Eq. (7.23) at 1 through 7:

v3 − 4v 2 + 5v1 = C v 4 − 4v3 + 6v 2 − 4v1 = C v5 − 4v4 + 6v3 − 4v2 + v1 = C v6 − 4v5 + 6v4 − 4v3 + v2 = 1C.5

(1) (2) (3) (4)

v7 − 4v6 + 6v5 − 4v 4 + v3 = C2 − 4v7 + 6v6 − 4v5 + v4 = C2 5v7 − 4v6 + v5 = C2

(5) (6) (7)

Solving, 4

Note:

pa vC = v 4 = 0.00966 pL EI = 0.1546 EI

( vC ) exact = 5 pa 4 32 EI

______________________________________________________________________________________ SOLUTION (7.9) po po/4 4

Let N = p0 L

-2

-1

A 0

po/2

h=L/4

3po/4

B 4

y Boundary conditions at A:

v ' (0) = 0; v1 = v −1 v (0) = 0; v0 = 0 v ' ' ' ' (0) = 0; v2 − 8v1 + v −2 = 0

Apply Eq. (7.23) at 1 through 4:

7v1 − 4v2 + v3 = (0.25)5 N

−4v1 + 6v2 − 4v3 + v4 = (0.5)9 N

v1 − 4v2 + 6v3 − 4v4 + v5 = 3(0.5)10 N v2 − 4v3 + 6v4 − 4v5 + v6 = (0.25) 4 N

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.9 (CONT.) Boundary conditions at B:

v ' ' ' ( L) = 0; v ' ' ( L) = 0;

− v 2 + 2v 3 − 2v5 + v 6 = 0 v 3 − 2v 4 + v5 = 0

The foregoing six equations are solved to yield

= v1 0.010742 = N v2 0.035156 N = v3 0.066406 = N v4 0.099609 N = v5 0.132812 = N v2 0.167969 N The error is therefore

0.099609 −0.091667 0.091667

× 100 = 8.7 % .

______________________________________________________________________________________ SOLUTION (7.10) 0

L/3

P/3

h=L/6

2P/3

Figure (a)

Application of Eq. (7.22) at points 1 through 5 gives, respectively:

v2 − 2v1 + v0 = − 9PLEI h 2

2 v3 − 2v2 + v1 = − 29 PL EI h 2 v4 − 2v3 + v2 = − PL EI h

v5 − 2v 4 + v3 = − PL EI h

(a)

2 v6 − 2v5 + v4 = − PL EI h

The boundary conditions are v0 = v6 = 0. Then Eqs. (a) may be represented

in matrix form as

0 0 0   v1  2  − 2 1  1 −2 1 0 0  v 2  4        1 −2 1 0   v3  =  3 C  0   0 1 − 2 1  v 4  2   0      0 0 0 1 − 2 v5  1  3

648EI . Solving the foregoing; v1 = −6.67C , v 2 = −11.33, v4 = −9.67C , v5 = −5.33C , and v3 = −12C , or

where C = − PL

v3 = 0.01852 PL EI

Note: The exact value of the deflection at the center (see Table D.4) is 0.01775PL EI . ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.11) p 1

-1

n=0

n=4

h=L/4

Figure (a)

We observe that because of symmetry only half of the beam span need be considered "

as shown in Fig. (a). From the boundary conditions v (0) = v (0) = 0 (Fig. 7.6) and symmetry:

v0 = v 4 = 0

v1 = −v −1

v1 = v3

By Eq. (7.23) at points 1 and 2,

(a)

v3 − 4v2 + 6v1 − 4v0 + v −1 = 0

(b)

v4 − 4v3 + 6v2 − 4v1 + v0 = phEI

Substituting Eqs. (a) into Eqs. (b), we obtain

6v1 − 4v 2 = 0

− 4v1 + 3v2 = phEI

Solving the above and setting h = L 4 yields v1 = 0.0039 pL

EI and

v2 = v max = 0.0059 pL EI

______________________________________________________________________________________ SOLUTION (7.12)

A y

0 1

P 4

h=a/2

P P

Pa x

Equation (7.22) is applied at nodes 1, 2, 3, respectively:

0 − 2v1 + v2 = 0; v1 − 2v 2 + v3 = 0;

v1 = v2 2 v 2 = 2v 3 3

v 2 − 2v3 + 0 = ( h 2 EI )( Pa 2) or

v3 = − 3Pah 2 8EI

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.12 (CONT.) Thus, and

v B = v 2 = − 161 PaEI = −0.0625 PaEI 3

v1 = − Pa 3 32 EI

Slop at A is then

θ A = v 2−hv = vh = − 16PaEI 1

−1

We have v B = 0.08333 Pa 0.08333−0.0625 0.08333

EI as exact solution. Error is thus × 100 = 25 %

______________________________________________________________________________________ SOLUTION (7.13) Boundary conditions yield

v (0) = 0; v ' (0) = 0;

and or or

v0 = 0 v1 = v −1

v ' ' ( L ) = 0 = v 4 − 2v 3 + v 2 v 4 = 2v 3 − v 2 v ' ' ' ( L) = 0 = v5 − 2v4 + 2v2 − v1

(1)

v5 = 4v3 − 4v2 + v1

(2)

Apply Eq. (7.23) at points 1, 2, and 3, with v0 = 0 and v1 = v −1 : 4

v3 − 4v 2 + 7v1 = 162pLEI

(3)

v 4 − 4v3 − 6v 2 − 4v1 = 81pLEI

(4)

v5 − 4v 4 + 6v3 − 4v 2 + v1 = 81pLEI

(5)

Solving Eqs. (1) to (6),

v3 = 7 pL4 54 EI

and

v1 = 4v3 21

v 2 = 4v 3 7

______________________________________________________________________________________ SOLUTION (7.14)

C = ph 4 EI

h=L 4

v1 = v −1 p

-1

A y

-3

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.14 (CONT.) Apply Eq. (7.22) at 1, 2, 3:

v3 − 4v 2 + 7v1 = C 5v3 − 4v2 + v1 = C

− 4v3 + v2 − 4v1 = C

ph 4

v3 = 1.36364 phEI

Solving, v1 = 0.90909 EI

v2 = v max = 1.68182 phEI

= 0.00657 pL EI ↓

θ B = θ max = 21h ( v5 − v3 ) = − vh

= −0.02131 pL EI

______________________________________________________________________________________ SOLUTION (7.15) P A B MA=M M 0 3 4 1 2 RA=P/2 RB=P/2 L/2 L/2 y PL/2 P PL/4 Mp x MR RA

-PL/8

x -PL/4

-3PL/8

-PL/2

v1 = v −1 , v1 = v3 (due to symmetry), v3 = v5 v0 = v4 = 0, C = L3 16 EI .

Apply Eq.(7.22) at 1through 4: o

v2 − 2v1 + v0 = ( − PL8 + M )C v3 − 2v2 + v1 = ( − PL4 + M )C

(1) (2)

v4 − 2v3 + v2 = ( PL4 − 3 PL 8 + M )C o

v5 − 2v 4 + v3 = ( PL4 − PL2 + M )C

(3) (4)

Equations (4), (3), (2) lead to

2v3 = MC v2 = ( 2 M − PL8 )C v1 = ( 92M − PL2 )C

Then, Eq. (1) gives

M = 81 PL Note: M exact = 81 PL

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.16) p L h=L/4 0

y Boundary conditions are v0 = v 4 = 0 and from symmetry v1 = v3 .

Then, using Eq. (7.23) at nodes 1 and 2: 4

v1 + 6v1 − 4v2 + v3 = phEI 4

Solving,

− 4v1 + 6v 2 − 4v3 = phEI 4

v 2 = phEI = EIP ( L4 ) 4 = 0.00391 pL EI

The exact solution is

v2 = 0.002604 pL EI

The slope is zero at nodes 0, 2, and 4. Thus,

θ= θ= 1 3

v2 − v0 2h

(0.00391)4 pL4 EI 2L 3

= 0.0078 pL EI

______________________________________________________________________________________ SOLUTION (7.17) F41

AE 13.5(10−4 )(69 ×109 ) = L 1.7 = 54.79 ×106 N m

1 150o L=1.7 m 4

F41

c= cos150o = − 3 2 s= sin150o = 12 (a)

Equation (7.38):

 34 3 4 − 3 4 −3 4   14 3 4 −1 4  6 − 3 4 [k ]e 54.79 ×10  =  34 − 3 4  −3 4 − 3 4   −1 4 − 3 4 14   3 4 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.17 (CONT.) or

(b)

 0.75 −0.433 −0.75 0.433   −0.433 0.25 0.433 −0.25   [k ]e = 54.79 MN m  −0.75 −0.433 0.75 −0.433    0.433 −0.25 −0.433 0.75 

Equation (7.39) with i=4 and j=1:

 3 = F41 54.79(106 )  −  2 or F1 = F41 = −73.13 kN (c)

1   2 + 1.1  −3  (10 )  2  1.5 + 1.2 

Then {δ }e = [T ]{δ }e , give

u4  v   4    u1   v1 

0.5 0 0   −1.1  −0.866  −0.5 −0.866 0 0  −1.2  =    0 0 0.5   2  −0.866   0 −0.5 −0.866   1.5   0

 0.35     1.59    mm 0.98 −    −2.3 

______________________________________________________________________________________ SOLUTION (7.18) We have

AE AE AE 2 AE AE 3 AE = ( )1 = ( )2 = ( )3 L L L L L L

Equation (7.25a):

= [k ]1

(a)

AE  1 −1 AE  2 −2  AE  3 −3 = = [k ]2 [k ]3     L  −1 1  L  −2 2  L  −3 3 

System stiffness matrix, [ k ] = [ k ]1 + [ k ]2 + [k ]3 is order of 4 × 4 . Thus

 1 −1 0 0    AE  −1 3 −2 0  [K ] = L  0 −2 5 −3    0 0 −3 3  (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.18 (CONT.) (b)

 F1x   1 −1 0 0   u1  F   −1 3 −2 0  u   2 x  AE    2   =  F3 x  L  0 −2 5 −3 u3     F4 x   0 0 −3 3  u4  Boundary conditions are u= 1

(1)

u= 0 . We have F3x = − P . Thus 4

 0  AE  3 −2  u2   =    − P  L  −2 5  u3  Solving

(c)

2 PL 3PL u2 = u3 = − − 11AE 11AE

Equations (1) :

 F1x   1 −1 0 0   0  F   −1 3 −2 0  − 2 P 11  2 x  AE   L   =   =   F3 x  L  0 −2 5 −3  − 3P 11 AE    0 0 −3 3   0   F4 x  The reactions are

2 R1 = P → 11

2 P 11  0       −P  9 P 11

9 R4 = P → 11

______________________________________________________________________________________ SOLUTION (7.19) We have

AE A(2 E ) 6 AE (= )1,2 = L L3 L AE 2 A( E ) 6 AE (= )3 = L L3 L

There are four displacement components ( u1 , u2 , u3 , u4 ) and so the order of the system matrix is 4x4. Using Eq. (7.25a):

 1 −1 [= k ]1 [k= ]2 [k= ]3 336(106 )    −1 1  (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.19 (CONT.)

(a)

−1 0 0 1  1 −1 0 0   −1 (1 + 1)  0  6 AE  −1 2 −1 0  −1 6 AE  [K ] = = −1 (1 + 1) −1 L 0 L  0 −1 2 −1     0 1 −1 0  0 0 −1 1  (b)

 F1x   1 −1 0 0   u1  F      2 x  AE  −1 2 −1 0  u2   =    F3 x  L  0 −1 2 −1 u3     F4 x   0 0 −1 1  u4  Boundary conditions are u= 1

(1)

u= 0 and Fx 3 = P . Equation (1) is then 4

 0  AE  2 −1 u2   =     P  L  −1 2  u3  Solving

= u2 (c)

PL PL = u3 9 AE 18 AE

Equations (1) result in

 F1x   1 −1 0 0   0  F   −1 2 −1 0   P 9   L  2 x  6 AE    =    = L  0 −1 2 −1  P 18 AE  F3 x     0 0 −1 1   0   F4 x  The reactions are

2 R1 =P ← 3

−2 P 3    P     0   − P 3 

1 R4 =P ← 3

______________________________________________________________________________________ SOLUTION (7.20)

AE AE ( = )1 L L Equation (7.25a):

[= ]2 k ]1 [k=

AE 4 AE AE ( = )3 = 0.8 L 5L L

AE AE ( = )2 L L

AE  1 −1 L  −1 1 

[= k ]3

AE  0.8 −0.8 L  −0.8 0.8 

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.20 (CONT.) (a) System matrix, [ K ] = [ k ]1 + [ k ]2 + [ k ]3 , is then

−1 0 0  1  −1 (1 + 1) −1 0  AE  [K ] = −1 (1 + 0.8) −0.8 L 0   0 −0.8 0.8  0 u1

(b)

0   u1   F1x   1 −1 0 F   0  u2  −1  2 x  AE  −1 2  =    F3 x  L  0 −1 1.8 −0.8 u3     F4 x   0 0 −0.8 0.8  u4  Boundary conditions are u= 1

Solving

u= 0 , We have Fx 3 = P . Thus 4

 0  AE  2 −1  u2   =     P  L  −1 1.8 u3 

u2 = (c)

(1)

PL PL u3 = 2.6 AE 1.3 AE

Equations (1) yield

0  0   F1x   1 −1 0 F   −1 2 0   P 2.6  L −1  2 x  AE  =   =    F3 x  L  0 −1 1.8 −0.8  P 1.3  AE    F4 x   0 0 −0.8 0.8   0  The reactions are

1 R1 = P ← 2.6

 − P 2.6    0     P   −1.6 P 2.6 

0.8 R4 = P ← 1.3

______________________________________________________________________________________ SOLUTION (7.21) Table P7.21 Data for the truss of Fig P7.21 Element Length(m) θ c s

7.5

36.9o

0.8

0.6

2 3 4

6 4.5 4.5

0 90o 0o

1 0 1

0 1 0

4.5 2

135o

−0.707 0.707

c2 cs s2 0.639 0.48 0.361 1 0 1

0 0 0

0 1 0

0.5

−0.5

0.5

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.21 (CONT.)

u1 v1 u2 v2  0.639 0.48 −0.639 −0.48   0.361 −0.48 −0.361 AE  0.48 [k ]1 = 0.639 0.48  7.5  −0.639 −0.48   0.361  −0.48 −0.361 0.48 u1 v1 u3 v3  1 0 −1 0   0 0  AE  0 0 [k ]2 = 1 0 6.0 − 1 0   0 0  0 0 u2 v 2 u3 v3 0 0 0 0 0 1 0 − 1  AE  [ k ]3 = 0 0 0 4.5 0   1 0 − 1 0 u3 v3 u4 v 4  1 0 −1 0   0 0  AE  0 0 [k ]4 = 1 0 4.5 − 1 0   0 0  0 0 u2 v2 u4 v4  0.5 −0.5 −0.5 0.5   AE  −0.5 0.5 0.5 −0.5 [k ]5 = 4.5 2  −0.5 0.5 0.5 −0.5    0.5 −0.5 −0.5 0.5

______________________________________________________________________________________ SOLUTION (7.22) Table P7.22 Data for the truss of Fig.P7.22

Element

Length (m)

1 2 3 4 5

2.4 2.6 1.0 2.4 2.6

0o 22.62o −90o 0o 22.62o

1 0 1 0 0 0.923 0.385 0.852 0.355 0.148 0 1 0 0 1 1 0 1 0 0 0.923 0.385 0.852 0.355 0.148

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.22 (CONT.) (a)

Use Eq.(7.38);

u A v A uC v C  1 0 −1 0   0 0  AE  0 0 [k ]1 = 1 0 2.4  −1 0   0 0  0 0 uA vA uB vB  0.852 0.355 −0.852 −0.355   AE  0.355 0.148 −0.355 −0.148 [k ]2 = 2.6  −0.852 −0.355 0.852 0.355    −0.355 −0.148 0.355 0.148 uB v B 0 0 0 1 AE  [k ]3 = 0 1.0 0  0 −1

uC v C 0 0 0 −1  0 0  0 1

uB v B  1 0  AE  0 0 [k ]4 = 2.4  −1 0   0 0

uD v D −1 0  0 0  1 0  0 0

uC vC uD vD  0.852 0.355 −0.852 −0.355   AE  0.355 0.148 −0.355 −0.148 [k ]5 = 2.6  −0.852 −0.355 0.852 0.355    −0.355 −0.148 0.355 0.148

(CONT.) ______________________________________________________________________________________ 203 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.22 (CONT.) (b)

Global Stiffness Matrix [ K ] =

uA vA uB vB uC vC uD vD 0.137 −0.328 −0.137 −0.417 0 0 0  0.745   0.137 −0.057 −0.137 −0.057  0 0 0 0    −0.328 −0.137  0.745 0.137 0 0 −0.833 0   −0.137 −0.057 0.137 1.057 0 −2 0 0   AE  −0.417 0 0 0 0.745 0.137 −0.328 −0.137    0 0 −2 0.137 1.057 −0.137 −0.057   0  0 0 −0.833 0 −0.655 −0.273 0.745 0.137    −0.137 −0.057 0 0 0 0.137 0.057   0

 RAx  0 R  0  Ay     0   uB      v B   0   = [K ]    uC   0  vC   0      uD   P   0   RDy   

{F } = [ K ]{δ };

______________________________________________________________________________________ SOLUTION (7.23) (a)

Element 1

θ= 135o c2 = 0.5 cs = −0.5 s2 = 0.5

Equation (7.38):

 0.5 −0.5 −0.5 0.5   AE  −0.5 0.5 0.5 −0.5 [k ]1 = L  −0.5 0.5 0.5 −0.5    0.5 −0.5 −0.5 0.5 Element 2 θ =

180o= c 2 1.0= cs 0= s2 0 u1 v1 u 3 v3

 1  AE  0 [k ]2 = L  −1   0

0 0 0 0

−1 0 1 0

0 0  0  0

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.23 (CONT.) Element 3 θ =

270o= c 2 0= cs 0= s 2 1.0 u1 v1 u4 v4

0 0 [k ]3 = k '  0  0 where k ' = kL AE (b)

0 1.0 0 −1.0

0 0 0 0

0  −1.0  0   1.0 

System matrix is 8x8. Superposition. [ K ] =

∑ [k ] results in u 3 v3 u 4

 1.5 −0.5 −0.5 0.5 −1  −0.5 0.5 0.5 −0.5 0   −0.5 0.5 0.5 −0.5 0  AE  0.5 −0.5 −0.5 0.5 0 [K ] = 0 0 0 1 L  −1  0 0 0 0  0  0 0 0 0 0  0 0 0  0 −k '

0 0 0 0 0 0 0 0

0 0 0 −k ' 0 0  0 0 0 0  0 0 0 0  k ' 0

where k ' = kL AE . (c)

Boundary conditions are

u= v= u= v= u= v= 0 2 2 3 3 4 4

Therefore

 0   u1  −P  u     2  F2 x  0      F2 y  0   = [K ]   F3 x  0  F3 y  0      F4 x  0  F4 y  0    

The total 7 × 7 stiffness matrix can be reduced to a 7 × 4 matrix by using the following antisymmetrical conditions: u1 = −u5 , u2 = −u4 , and v2 = v1

______________________________________________________________________________________ 205 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (7.24) Table P7.24 Data for the truss of Fig.P7.24 Element Length(m) θ c s

1 5 2 4 ( a ) Use Eq.(7.38):

53.13o 180o

0.6 0.8 0.36 0.48 0.64 −1 0 1 0 0

u1 v1 u2 v2 0.48 − 0.36 − 0.48  0.36  0.48 0.64 − 0.48 − 0.64 AE  [k ]1 = 0.36 0.48 5 − 0.36 − 0.48   0.48 0.64 − 0.48 − 0.64 u2 v 2 u3 v3  1 0 −1 0   0 0  AE  0 0 [k ]2 = 1 0 4 − 1 0   0 0  0 0

(b)

(c)

Global Stiffness Matrix

u1 v1 u2 v2 u3 v3 0.096 − 0.072 − 0.096 0 0  0.072  0.096 0.128 − 0.096 − 0.128 0 0   − 0.072 − 0.096 0.322 0.096 − 0.25 0  [ K ] = AE   0.096 0.128 0 0 − 0.096 − 0.128  0 0 − 0.25 0 0.25 0    0 0 0 0 0   0

 0  0.322 0.096  u2    = AE    −6000  0.096 0.128  v2 

u2    0   0.9  −3 1  4 −3 =   =     (10 ) m  6   v2  20(10 )  −3 10.063 −6000  −3.02  (d)

 R1x   −0.072 −0.096   4.5024  u2        AE  −0.096 −0.128    =  6.0032  kN  R1 y  =  v2  −4.5    0    R3 x   −0.25 

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.24 (CONT.) ( e ) Use Eq.(7.39)

u   0.0018  F12 = AE [ 0.6 0.8]  2  = 20(106 ) [ 0.6 0.8]   = −75 kN (C ) −0.00604   v2  −u   −0.0018  20(106 ) [ −1 0]  36 kN (T ) F23 = AE [ −1 0]  2  = =  0.00604   −v2 

______________________________________________________________________________________ SOLUTION (7.25) We have E=

105 GPa A= 10 ×10−4 m 2

Table P7.25 Data for the truss of Fig.P7.25 Element Length(m) c s θ

1 2

53.13 o 90 o

5 4

c2 cs s2 0.6 0.8 0.36 0.48 0.64 0 1 0 0 1

( a ) Apply Eq.(7.38):

u1 v1 u2 v2 0.48 − 0.36 − 0.48  0.36  0.48 0.64 − 0.48 − 0.64 AE  [k ]1 = 0.36 0.48 5 − 0.36 − 0.48   0.48 0.64 − 0.48 − 0.64

u1 0  AE 0 [k ]2 = 4 0  0

v1 0

u3 0

1 0 −1

0 0 0

v3 0 − 1  0  1

( b ) Global Stiffness Matrix

u1 v1 u2 v2 u3 1.008 − 0.756 − 1.008 0  0.756  1.008 3.969 − 1.008 − 1.344 0  − 0.756 − 1.008 0.756 1.008 0 [ K ] = 10 7  1.008 1.344 0  − 1.008 − 1.344  0 0 0 0 0  − 2.625 0 0 0  0

v3 0

 − 2.625  0  0   0  2.625

(CONT.) ______________________________________________________________________________________ 207 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.25 (CONT.)

1.008  −0.015  F1x  7  0.756  = 10    10, 000  1.008 3.969   v1  3 10 ×10= (1.008 ×107 )(−0.015) + 3.969 ×107 v1

( c )

F1x = (0.756 ×10 )(−0.015) + 1.008 ×10 v1 (d)

Support reactions:

 F2 x   −0.756 −1.008  F   −1.344  −0.015   2y  7  −1.008 = =    10   0   0.0041 0  F3 x     F3 y  −2.625   0

or or

v1 = 0.0041 m

F1x = −72.1 kN

 72.1   96.1      kN 0   −107.6 

______________________________________________________________________________________ SOLUTION (7.26) We have AE = 20 MN. Table P7.26 Data for the truss of Fig.P7.26 Element Length θ c s c2 cs s2

1 2 3

2.5 4 2,5

36.87 o 0o 143.13 o

0.8 0.6 0.64 0.48 0.36 1 0 1 0 0 − 0.8 0.6 0.64 − 0.48 0.36

( a ) Apply Eq.(7.38):

u1 v1 u2 v2 0.48 − 0.64 − 0.48 u1  0.64  0.48 0.36 − 0.48 − 0.36 v1 AE  [k ]1 = 0.64 0.48 u 2 2.5 − 0.64 − 0.48   0.48 0.36 v 2 − 0.48 − 0.36

u1  1  AE  0 [k ] 2 = 4 − 1   0

v1 u3 v3 0 − 1 0  u1 0 0 0  v1 0 1 0  u3  0 0 0  v3

u2 v2 u3 v3 0.48 u 2  0.64 − 0.48 − 0.64 − 0.48 0.36 0.48 − 0.36 v 2 AE  [ k ]3 = 0.48 0.64 − 0.48 u 3 2.5 − 0.64   0.36 v3  0.48 − 0.36 − 0.48

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.26 (CONT.) ( b ) Global Stiffness Matrix

u1 v1 u2 v2 u3 v3 0.192 − 0.256 − 0.192 − 0.250 0  u1  0.506 v  0.192 0.144 − 0.192 − 0.144 0 0  1  − 0.256 − 0.192 0.512 0 0.192 u 2 − 0.256 [ K ] = AE   0 0.288 0.192 − 0.144 v 2 − 0.192 − 0.144 − 0.250 0 0.192 0.506 − 0.192 u 3 − 0.256   0 0.192 − 0.144 − 0.192 0.144 v3  0

−0.256  u2   F2 x   0.512 0      ( c )  F2 y  = AE 0 0.288 0.192  v2   F   −0.256 0.192 0.506  u3   3x 

{δ } = AE1 [ K ]−1 { F } u2   2.9531 −1.3333 2.0   30000   0.009096         1 −2.6667  −70000  = 1.3333 5.25  v2  =− −0.020375  m 20×106      0   0.012333  −2.6667 4.0 u3   2.0  F1x   −0.256 −0.192 −0.25   0.009096  −29.997        −0.020375  = AE  −0.192 −0.144 0. ( d )  F1 y  =   23.751  kN         F3 x   0.192 −0.144 −0.192   0.012333   46.250  ( e ) Use Eq.(7.39): We have u1 = v1 = v3 = 0.

F12 =

= F13

 0.009096  AE [0.8 0.6] −0.020375 = −39.59 kN (C ) 2.5   0.012333 AE = [1 0]  0  61.67 kN (T ) 4  

0.009096 − 0.012333 AE F32 = −77.08 kN (C ) [ − 0.8 0.6]  −0.020375  = 2.5   ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.27) Equation (7.38):

 12 6 L −12 6 L   2 2 EI  6 L 4 L −6 L 2 L  [= k ] [K ] = 12 −6 L  L3  −12 −6 L  2 2  6 L 2 L −6 L 4 L 

(a)

(b)

Boundary conditions are v= 2

θ= 0 . Equation (7.45a) becomes 2

− P  EI  12 6 L   v1   = 2  2   0  L 6 L 4 L  θ1  PL2 PL3 Solving, θ1 = v1 = − 2 EI 3EI

Equation (7.45a):

Solving

where

 F1 y   12 6 L −12 6 L  − PL3 3EI   M   2 2 2  1  EI  6 L 4 L −6 L 2 L   PL 2 EI    = 3  12 −6 L   0   F2 y  L  −12 −6 L  2 2  M 2   0  6 L 2 L −6 L 4 L  

 F1 y   − P  M   0   1    =   F2 y   P   M 2  − PL  F2= R= P. y 2

______________________________________________________________________________________ SOLUTION (7.28) Due to the symmetry only one-half of the beam need be considered. Referring to Case 3 of Table D.5, equivalent nodal forces obtained (see Fig. a). pL/4 pL2/48

Boundary conditions are v= 1

pL/4

 pL2 48 EI  4 L12  = 3   − pL 4  L1  −6 L1

48 and F2 y = − pL 4

We have M 1 = − pL

2 2 pL /48 L L1 = 2 Figure (a) Equivalent nodal forces

θ= 0 2

Equation (7.45a), with element of length L1 = L 2 .

−6 L1  θ1  8 EI  L12   = 3  L  −3L1 12  v2 

−3L  θ1    12  v2 

(CONT.) ______________________________________________________________________________________ 210 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.28 (CONT.) Inverting

θ1  L3 12 L2 3 L   pL2 48  =    1   − pL 4  v2  24 EI  3 L θ1   − pL3 24 EI   =  4 v2  −5 pL 384 EI 

This is the exact solution (See Table D.4) ______________________________________________________________________________________ SOLUTION (7.29) Due to symmetry, only one-half of the beam need be considered.

 12 6 L −12 6 L   2 2 EI  6 L 4 L −6 L 2 L  [k ]1 = 3 12 −6 L  L  −12 −6 L  2 2  6 L 2 L −6 L 4 L 

P/2

1 1

k/2

 0  3 EI  kL EI [k ]2 = 3 L  0   0

0 0 0 0

0 0  0  0

Therefore

−12 6 L   12 6 L  6 L 4 L2 −6 L 2 L2   EI [K ] = 3   kL3 L  −12 −6 L 12 + −6 L  EI   2 −6 L 4 L2   6 L 2 L

(a)

Boundary conditions are v1 = 0 and θ 2 = 0. Equation (7.45a) with F2 y = − P 2 and M 1 = 0 :

 4 L2 −6 L  θ1   0  EI  3  = 3  kL  v2  − P 2  L −6 L 12 +  EI  Introduce the data and solve:

v2 = −7.9338 mm θ1 = −2.9752 rad

(CONT.) ______________________________________________________________________________________ 211 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.29 (CONT.) (b)

−12 6 L   12 6 L  F1 y   0   6 L 4 L2 2 − L L 6 2  0    −2.9752  −3   EI  3   (10 )  = 3  kL −6 L   −7.9338  F2 y  L  −12 −6 L 12 + EI   0  2   M 2  −6 L 4 L2    6 L 2 L

 F1 y     F2 y =  M   2

 5.20 kN     −3.80 kN  −20.8 kN ⋅ m   

= Fspring 200(7.9338) = 1.587 kN (C ) From symmetry: F= 1y

F= 5.20 kN ↑ 3y

______________________________________________________________________________________ SOLUTION (7.30) P/2 PL/8

P/4

Figure (a). Equivalent Nodal forces. Equation (7.45a) reduces to

PL/8

Boundary conditions are

= θ1 0 v1 0=

We have

PL M2 = −M1 = 8 P F1 y = F2 y = − 2

− P 2  EI  12 −6 L   v1   = 3   2  PL 8  L  −6 L 4 L  θ 2 

Inverting

 v2  L  L2 3L  − P 2  −5 PL3 48 EI   = =    2 θ 2  6 EI 3L 6   PL 8   − PL 8 EI 

______________________________________________________________________________________ SOLUTION (7.31)

EI EI 4 EI (= )2 = L L3 4 L3 So, the stiffness materices [ k ]1 + [ k ]2 and [ K ] are the same as obtained in Example 7.9.

We have (

EI EI )1 = 3 L L

and

Boundary conditions are= v1

becomes

0,= θ1 0, and v3 = 0. Equation below Eq.(e) of Example 7.9

(CONT.) ______________________________________________________________________________________ 212 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.31 (CONT.)

18 −30   − P   v2   28 L L2      18 51 L − 39 L  23PL  = θ 2     276 EI  −30 − 39 L 111 L   0  θ3      8   −2  3 PL3   PL   = =  84 L   21 L  276 EI   69 EI −12 L  − 48 L    Substituting the given data:

 v2   −8  3 3   (30 ×10 )(1.2)   = θ 2  = 168  3 θ  69(207 ×10 )(15) −96   3  

−1.936 mm     0.041 rad   −0.023 rad   

______________________________________________________________________________________ SOLUTION (7.32) (a)

The three element stiffness are identical. Equation (7.46):

 12 6 L −12 6 L   2 2 EI  6 L 4 L −6 L 2 L  [= k ]1 [k= ]2 [k= ]3 12 −6 L  L3  −12 −6 L  2 2  6 L 2 L −6 L 4 L 

(b) The beam stiffness matrix, [ K ] = [ k ]1 + [ k ]2 + [ k ]3 , is assembled as

θ1

6L  12  6L 4 L2   −12 −6 L  2 L2 EI  6 L [K ] = 3 0 L  0  0  0  0 0  0  0

θ2

−12 6L −6 L 2 L2 24 0 0 8 L2 −12 L −6 L 6L 2 L2 0 0 0 0

θ3

0 0 0 0 −12 L 6L −6 L 2 L2 24 0 0 8 L2 −12 −6 L 6L 2 L2

θ4

0 0  0 0  0 0   0 0  6L  −12  2 L2  −6 L 12 −6 L   4 L2  −6 L

Boundary conditions are :

v= v= v= θ= 0 2 3 4 4

System governing relations, from . (7.40): (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.32 (CONT.)

 −P  v1   0 θ     1  Fy 2  0     M 2  θ 2    = [K ]    Fy 3  0 M3  θ3       Fy 4  0 M    0  4

______________________________________________________________________________________ SOLUTION (7.33) (a) Use Eq.(7.46):

v1 θ1 v2 θ2  12 6 L −12 6 L   2 2 EI 6 L 4 L −6 L 2 L  [k ]1 = 3  12 −6 L  L  −12 −6 L  2 2  6 L 2 L −6 L 4 L  v2 θ2 v3 θ3

 12 6 L −12 6 L   2 2 EI  6 L 4 L −6 L 2 L  [k ]2 = 3 12 −6 L  L  −12 −6 L  2 2  6 L 2 L −6 L 4 L  (b)

Assemble the global stiffness matrix of the beam: [ K = ]

[k ]1 + [k ]2 . Then,

6 L −12 6 L 0 0   v1   R1   12 M   6 L 4 L2 −6 L 2 L2 0 0  θ1   1   R2  EI  −12 −6 L 24 0 −12 6 L   v2   = 3    2 2 0 8 L −6 L 2 L2  θ 2  M 2  L  6L 2L  F3 y   0 0 −12 −6 L 12 −6 L   v2       0 6 L 2 L2 −6 L 4 L2  θ3   M 3   0 (c) The boundary conditions are= v1

(1)

0,= θ1 0, and v2 = 0 , Hence

− P   12 6 L   EI  2  0  = 3 6 L 4 L L 0  6 L 2 L2  

6 L   v3    2 L2  θ3  8 L2  θ 2 

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.33 (CONT.) Solving

7 PL3 3PL2 PL2 , , v3 = − θ3 = θ2 = 12 EI 4 EI 4 EI

(d)

Introducing these equations into Eq. (1), after multiplying :

(e)

P ki

5 0, F3 y = − P, M 3 == R2 P 2 3 1 M2 = − PL, R1 = − P, M1 = PL 2 2

3 P

-3P/2 M PL/2

x PL

______________________________________________________________________________________ SOLUTION (7.34)

= L 7 m, P = 8 kN, = EI = 65(105 ) N ⋅ m 2 , k 200 kN m v1 θ1 u2 θ4 θ 2 u3 θ3 v2 [k ]1

 12 6 L −12 6 L   0  6 L 4 L2 −6 L 2 L2   3 EI  EI  kL EI  [k ]2 = 12 −6 L  L3  −12 −6 L L3  0   2 2  6 L 2 L −6 L 4 L   0

0 0 0 0

0 0  0  0

Thus

−12 6 L   12 6 L  6 L 4 L2 −6 L 2 L2   EI [K ] = 3   kL3 L  −12 −6 L 12 + −6 L  EI   2 −6 L 4 L2   6 L 2 L

(1)

(CONT.) ______________________________________________________________________________________ 215 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.34 (CONT.) ( a ) Boundary conditions are v1 = 0 and θ1 = 0 . Equation (7.45a), with F2 y = − P and M 2 = 0 :

  kL3 − P  EI 12 + −6 L   v2  EI  = 3    2 θ 2  0  L  −6 L 4 L   −1 1 {δ } = AE [ K ] { F }

Substituting the given numerical values and solving, we have

v2 = −0.031146 m θ 2 = −0.006674 rad

(b)

−12 6 L   12 6 L 0  F1 y     6 L 4 L2 2 6 2 L L − M     0   1  EI    = 3   kL3 −6 L  −0.031146   F2 y  L  −12 −6 L 12 + EI   −0.006674  2   M 2  −6 L 4 L2    6 L 2 L

Introducing the data and multiplying:

1.771 kN  F1 y     M   12.395 kN ⋅ m   1    =  −8.0 kN  F2 y     M 2  −0.000531 kN ⋅ m  The spring force is Pspring 200(0.031146) = = 6.229 kN

(C )

______________________________________________________________________________________ SOLUTION (7.35) Table P7.35 Data for the truss of Fig.P7.35

θ 0o

2 3 4 5

60 o 120 o 0o 60 o

0.5 0.866 0.25 0.433 0.75 − 0.5 0.866 0.25 − 0.433 0.75 1 0 1 0 0 0.5 0.866 0.25 0.433 0.75

u1  1  AE  0 [k ]1 = L − 1   0

v1 0

0 0 0

cs 0

s2 0

c 1

( a ) Use Eq.(7.49):

s 0

c2 1

Element 1

u4 v 4 − 1 0  u1 0 0  v1  1 0  u4  0 0  v4

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.35 (CONT.)

u1 v1 u2 v2 0.433 − 0.25 − 0.433 u1  0.25  0.433 0.75 − 0.433 − 0.75  v1 AE   [k ]2 = 0.25 0.433 u2 L − 0.25 − 0.433   0.433 0.75  v 2 − 0.433 − 0.75

u2 v2 u4 v4 0.433 u2  0.25 − 0.433 − 0.25 − 0.433 0.75 0.433 − 0.75  v 2 AE   [k ]3 = 0.433 0.25 − 0.433 u4 L − 0.25   0.75  v 4  0.433 − 0.75 − 0.433 u2  1  AE  0 [k ]4 = L − 1   0

v2 u3 v3 0 − 1 0  u2 0 0 0  v2  0 1 0  u3  0 0 0  v3

u3 v3 u4 v4 0.433 − 0.25 − 0.433 u3  0.25  0.433 0.75 − 0.433 − 0.75  v3 AE   [k ]5 = 0.25 0.433 u4 L − 0.25 − 0.433   0.433 0.75  v 4 − 0.433 − 0.75 ( b ) Global Stiffness Matrix [ K ] =

u1 v1 . 0.433  125  0.433 0.75  − 0.25 − 0.433  AE − 0.433 − 0.75 0 L  0  0  0 − 1 0  0  0

u2 − 0.25

v2 − 0.433

u3 0

v3 0

u4 −1

v4 0

 u1  v − 0.433 − 0.75 0 0 0 0  1 15 . 0 −1 0 − 0.25 0.433 u2  0 15 . 0 0 0.433 − 0.75  v 2 −1 0 125 . 0.433 − 0.25 − 0.433 u3  0 0 0.433 0.75 − 0.433 − 0.75  v3 u − 0.25 0.433 − 0.25 − 0.433 15 . 0  4 .  v 4 0.433 − 0.75 − 0.433 − 0.75 0 15

(CONT.) ______________________________________________________________________________________ 217 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.35 (CONT.)

{F } = [ K ]{δ };

 R1x  0 R  0  1y     0  u2      v 2   P    = [K ]    u3  Q  v3   0      0  R4 x   0   R4 y   

______________________________________________________________________________________ SOLUTION (7.36) We have AE=60 MN. Table P7.36 Data for the truss of Fig.P7.36

Element 1 2

θ Length 5 5313 . o 4 0o

c s c2 cs s2 0.6 0.8 0.36 0.48 0.64 1 0 1 0 0

( a ) Use Eq.(7.38):

u1 v1 u2 v2 0.48 − 0.36 − 0.48 u1  0.36  0.48 0.64 − 0.48 − 0.64 v1 AE   [k ]1 = 0.36 0.48 u2 5 − 0.36 − 0.48   0.48 0.64 v 2 − 0.48 − 0.64 u2 v 2 u3 v3  1 0 − 1 0  u2  0 0  v2 AE  0 0  [k ]2 = 1 0  u3 4 − 1 0   0 0  v3  0 0

(CONT.) ______________________________________________________________________________________ 218 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ (CONT.) ( b ) Global Stiffness Matrix

u1 v1 u2 v2 u3 v3 0.096 − 0.072 − 0.096 0 0  u1  0.072  0.096 . − 0.096 − 0128 . 0128 0 0  v1   − 0.072 − 0.096 0.322 0.096 − 0.25 0  u2 [ K ] = AE   . . 0.096 0128 0 0  v2 − 0.096 − 0128  0 − 0.250 0 0 0.25 0  u3   0 0 0 0 0  v3  0

( c ) Boundary Conditions: u1 = v1 = u3 = v3 = 0.

 F2 x  0.322 0.096 u2    = AE    .  v 2  0.096 0128  F2 y  −3   0   0.0005  u2  1  4 =   =    m    v2  AE  −3 10.063 −10, 000  −0.002   F1x   −0.072 −0.096  F   −0.096 −0.128 0.0005   1y     (d)  = =   AE  −0.250 0  −0.002   F3 x     F3 y  0   0

 9360    12480   N −7500   0 

( e ) Use Eq.(7.39)

F12 =

 0.0005  AE [0.6 0.8] −0.002 = −15.6 kN (C ) 5  

F23 =

−0.0005 AE [1 0]  0.002  = −7.5 kN (C ) 4  

______________________________________________________________________________________ SOLUTIONS (7.37 through 7.40) It is important to take into account any condition of symmetry which may exist. Use a twodimensional finite element program or ANSYS. ______________________________________________________________________________________ SOLUTION (7.41) ( a ) Refer to Fig. 7.25a: p −p

( 2 p j + pm ) 6

p −p

( 2 pm + p j ) 6

Q j = 12 p j h1t + 13 m 2 j h1t = h1t Qm = 12 p j h1t + 23 m 2 j h1t = h1t

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.41 (CONT.) ( b ) Refer to Fig. 7.25a:

2 2 3P τ xy = PQ It = 4 th ( h − y )

(a)

Substituting Eq. (a) into the given expression for Qm :

Qm = ym 1− y j ∫

3P 3 y j 4h

( h 2 − y 2 )( y − y j )dy 3

= ym 1− y j 34Ph [ − y j ( y − 3yh 2 )

ym yj

+ ( y2 − 4yh 2 )

ym yj

]

This leads to the first of Eqs. (7.79). Similarly, ym

Q j = ∫ τ xy tdy − Qm

(b)

Substituting Eq. (a) and Qm (from the first of Eqs. 7.79), Eq. (b) leads to the second of Eqs. (7.79).

______________________________________________________________________________________ SOLUTION (7.42) Qy3 3 1 cm

(b)

1 cm

(a)

4 cm

Qy4 4 x P=5 kN 2

Body force effects:

{Q}ba = 13 At{Q x1 , Q x 2 , Q x 3 , Q y1 , Q y 2 , Q y 3 } 1 = 3 (4 × 0.3){0, 0, 0, − 0.077, − 0.077, − 0.077} N

b {Q}= {0, 0, 0, 0, − 0.0308, − 0.0308, − 0.0308, − 0.0308, 0} N a

Similarly,

Q}bb {0, 0, 0, 0, 0, − 0.0308, − 0.0308, − 0.0308} N {=

Hence, or

{Q}b = {Q}ba + {Q}bb b {Q}= {0, 0, 0, 0, − 0.0308, − 0.0616, − 0.0616, − 0.0308} N

Effect of shear force, P: From Case Study 7.1,

{Q= }P {0, 0, 0, 0, − 2500, − 2500} N

Surface traction effects, p: Total load (4 × 0.3)700 = 840 N is equally divided between nodes 3 and 4. Thus,

= {Q}P {0, 0, 0, 0, 0, 0, − 420, − 420} N (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.42 (CONT.) Thermal strain effects: We have= ε0

Hence,

= 600 µ and 12(10−6 )50

 b1 1  [ B ]a = 0 2A  a1

b2 0 a2

b3 0 a3

0 a1 b1

0 a2 b2

0 a3   b3 

− 2 1 = 0 8 − 4

0 0

0 4

−4 −2

0 2

0 4  0 

{Q}ta = [ B ]Ta [ D ]{ε 0 }( At )

 −2  2  1 0  8 0  0   0

0 0 0 −4 0 4

−4  0  0.3 0  600 µ  1 4  2(107 )    0.3 1 0  600 µ  (1.2)   −2  0.91 0 0 0.35  0  2  0

After multiplication, this gives

{Q}ta = {−5142.85, 5142.85, 0, − 10285.7, 0, 10285.7} N

{Q}ta = {−5142.85, 5142.85, 0, 0, − 10285.7, 0, 10285.7, 0} N

Similarly,

Thus,

 b2 1  [ B ]b = 0 2A  a 2

b4 0

b3 0

0 a2

0 a4

 0 1 = 0 8 − 4

2 0 4

−2 0 0

0 −4 0

0 4 2

0 a3   b3  0 0  − 2 

{Q}t = [ B ]Ta [ D ]{ε 0 }( At ) = {0, 5142.85, − 5142.85, − 10285.7, 10285.7, 0} N {Q}tb = {0, 0, − 5142.85, 5142.85, 0, − 10285.7, 0, 10285.7} N

(CONT.) ______________________________________________________________________________________ 221 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 7.42 (CONT.) Hence, or

{Q}t = {Q}ta + {Q}tb

{Q}t = {−5142.85, 5142.85, − 5142.85, 5142.85, − 10285.7, − 10285.7, 10285.7, 10285.7} N

The system nodal force matrix:

{Q} = {Q}b + {Q}P + {Q} p + {Q}t = {−5142.85, 5142.85, − 5142.85, 5142.85, − 10285.7, − 12785.76, 986.64, 7365.67} N

System equation:

{Q} = [ K ]{0, u2 , 0, u4 , 0, v2 , 0, v4 }

where

[K ] is given by Eq. (d) of Case Study 7.2.

Since we have only 4 unknown quantities u2 , u4 , v 2 , v 4 (and 8 equations are available), there are redundant equations. Examination of these system of equations shows that:

[K ] is reduced by crossing out; row 1 and column 1 for u1 = 0, row 3 and column 3 for u3 = 0, row 5 and column 5 for v1 = 0, row 7 and column 7 for v3 = 0. {Q} is reduced by crossing out Q x1 , Q x 3 , Q y1 , Q y 3 . for

u1 = u3 = v1 = v3 = 0 Thus, from the reduced equations, we obtain

0.180 0.252 0.247  5142.85  u2  0.429 u  0.180 0.483 − 0.256 − 0.351  5142.85   4 −6      = 10 v − 0 . 252 0 . 256 1 . 366 1 . 373   − 12785.76   2    v 4  1.373 1.546  7365.67  0.247 − 0.351 or

u2   1729  u   4098   4   −6  = 10 cm  v2  −7373   v4  −6702 

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (7.43) y 3 0 1

4 (a)

1 cm

(b) 4 cm

1 cm

Stiffness matrix of element a: Let i = 1, j = 4, m = 3. Then,

x1 = 0

x4 = 4

x3 = 0

y1 = −1

y4 = 4

y3 = 1

Equations (7.68) give

We have

a1 = 0 − 4 = −4 a4 = 0 − 0 = 0 a3 = 4 − 0 = 4

b1 = 1 − 1 = 0 b4 = 1 + 1 = 2 b3 = −1 − 1 = −2

3.33 0.99 0  10 6  [D ] = 0.99 3.3 0    8 0 1.16 0 6 Equations (7.76) in 10 are thus, k uu ,11 = [0 + 1.16(16)] 8 = 2.32 *

k uu ,14 = [0 + 0 + 0] 8 = 0

k uu ,13 = [0 + 1.16( −16) + 0] 8 = −2.32

k uu , 44 = [3.3( 4) + 0 + 0] 8 = 1.65

k uu , 43 = [3.3( −4) + 0 + 0] 8 = −1.65 kuu ,33= [3.3(4) + 1.16(16) + 0] 8= 3.97 These can be written as follows:

0   2.32 − 2.32  k uu = − 2.32 3.97 − 1.65(106 )   1.65 − 1.65  0 Similarly, we find submatrices k vv and k uv . In so doing and after assembling these matrices, 6

we obtain the stiffness matrix for the element a (in 10 ):

0 0 1.16 − 1.16  2.32 − 2.32 − 2.32 3.97 − 1.65 0.99 − 2.15 1.16   − 1.65 1.65 − 0.99 0.99 0   0 [ k ]a =   0.99 − 0.99 6.6 − 6.6 0   0  1.16 − 2.15 0.99 − 6.6 7.18 − 0.58   1.16 0 0 − 0.58 0.58  − 1.16

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.43 (CONT.) Stiffness matrix for element b: Let i = 1, j = 2, m = 4. Then,

a1 = 4 − 4 = 0 a 2 = 0 − 4 = −4 a4 = 4 − 0 = 4

b1 = −1 − 1 = −2 b2 = 1 + 1 = 2 b4 = −1 + 1 = 0 6

Then, Eqs. (7.54) yield (in 10 ):

k uu ,11 = [3.3( 4) + 0 + 0] 8 = 1.65

k uu ,12 = [3.3( −4) + 0 + 0] 8 = −1.65 k uu ,14 = [0 + 0 + 0] 8 = 0 k uu , 22 = [3.3( 4) + 1.16(16) + 0] 8 = 3.97

k uu , 24 = [0 + 1.16( −16) + 0] 8 = −2.32 k uu , 44 = [0 + 1.16(16) + 0] 8 = 2.32 or

0   1.65 − 1.65  3.97 − 2.32(10 6 ) k uu = − 1.65   2.32 − 2.32  0

Similarly, we obtain submatrices k vv and k uv . In so doing and after assembling these matrices, 6

we determine the stiffness matrix for the element b (in 10 ):

0 0 0.99 − 0.99  1.65 − 1.65  − 1.65 3.97 − 2.32 1.16 − 2.15 0.99   − 2.32 2.32 − 1.16 1.16 0   0 [k ]b =   1.16 − 1.16 0.58 − 0.58 0   0  0.99 − 2.15 1.16 − 0.58 7.18 − 6.6    0.99 0 0 − 6.6 6.6  − 0.99 Prior to addition: the 2nd and 6th rows and columns of zeros are added to the matrix [k ]a ; the 3rd and 7th rows and columns of zeros are added to the matrix [k ]b . The system matrix:

[ K ] = [k ]a + [k ]b (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.43 (CONT.) 6

is then (in 10 ):

0 0 0.99 1.16 − 2.15  3.97 − 1.65 − 2.32  − 1.65 3.97 0 − 2.32 1.16 − 2.15 0 0.99   0 3.97 − 1.65 0.99 0 − 2.15 1.16 − 2.32   0 − 2.32 − 1.65 3.97 − 2.15 1.16 0.99 0   [K ] =  0 1.16 0.99 − 2.15 7.18 − 0.58 − 6.6 0    0 1.16 − 0.58 7.18 0 − 6.6   0.99 − 2.15  1.66 0 − 2.15 0.99 − 6.6 0 7.18 − 0.58   0.99 1.16 0 0 − 6.6 − 0.58 7.18 − 2.15 The force-displacement relation, Eq. (d) of Case Study 7.2 becomes

 0   3.97 −2.32 −2.15 0.99  u2   0   −2.32 3.97 1.16 0  u       4   = −2.5  −2.15 1.16 7.18 −6.6   v2    7.18   v4  −6.6 −2.5  0.99 0

This yields

179.9 357.9 225.8  0  u2  483.2 u  178.6 − 246.8 429 251.7  0   4 −6     = 10  v − − 357 . 9 247 1546 1373  − 2.5   2   v 4  − 1366  − 2.5 255.8 − 251.7 − 1373 −1459.25  −12.25   −6 =  (10 ) m 432.5 −    6849.5 

Stresses in element b:

εx   b1 1    0 ε y  = 2A  γ  a1  xy  b

b2 0 a2

b4 0 a4

0 a1 b1

0 a2 b2

 u1  u  0  2  u  a4   4   v b4   1  v2    v4 

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 7.43 (CONT.)

 −2 10−6  0 = 8   0

2 0

0 0

0 −4

−4

−2

= {−364.81, 3640, 615.37} µ

0   −1459.25  0    12.25 − 4    0   0   −432.5     6847.5 

Thus,

σ x  0.3 0  −364.81  1   200(103 )    0.3 1 0   3640.  σ y  =  0.91 τ  0 0 0.35  615.37  xy  b = {159.82, 775.95, − 47.34} MPa Stresses in element a:

εx   0   10−6  0 = ε y  8  γ   −4  xy a

−2 0

2 0

0 −4

0 4

−2

= {−1.53, 0, 1712.38} µ

 0   0   0    − 12.25 0    0   2   0     6849.5 

σ x  0.3 0   −1.53 1 200(103 )      0.3 1 0   0  σ y  =  0.91 τ  0 0 0.35 1712.38  xy a = {0.34, − 0.1, 93.62} MPa End of Chapter 7

CHAPTER 8 SOLUTION (8.1) ( a ) From Eq. (8.13), we have 2

σ θ ,min = ba −pa (1 + bb ) = pi b2−aa i

σ θ ,max = ba −pa (1 + ab ) = pi ba −+ab i

Hence,

σ θ , max σ θ , min

= a2+a b2 = a +2(1a.21a ) = 1.105 2

( b ) Using Eq. (8.16),

σ θ ,max = −2 p0 b b−a 2

σ θ ,min = − p Thus, σ θ , max σ θ , min

a 2 + b2 0 b2 − a 2 2

= b22−ba 2 = 2 (21..2121aa2) = 1.1 2

______________________________________________________________________________________ SOLUTION (8.2) Equation (8.20): 2

σ z = bp−aa = 10.−60.p6 = 0.5625 pi = 140 i

i 2

pi = 248.9 MPa

Equation (8.13):

= σ θ ,max or

b2 + a 2 b2 − a 2

12 + 0.62 12 − 0.62

pi 2.125 pi 140 = =

pi =

pi = 65.9 MPa

Equation (8.10):

τ max = bp−ba = 1 −10.6 pi = 1.5625 pi = 80 i

= pi 51.2 = MPa pall

______________________________________________________________________________________ SOLUTION (8.3) ( a ) Initial maximum tangential stress, from Eq. (8.13),

σ θ = pi bb +−aa = nn +−11 pi

pi = σ

n 2 −1 θ n 2 +1

After boring, denoting the inner radius by rx , we have 2 2

2 2

∆σ θ + σ θ = pi nn 2aa 2 +− rrx2 = nn 2 +−11 σ θ nn 2aa 2 +− rrx2 2

( ∆σ θ + σ θ )( n + 1)( n a − r ) = ( n 2 − 1)σ θ ( n 2 a 2 + rx2 ) 2

2 x

(CONT.) ______________________________________________________________________________________ 227 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 8.3 (CONT.) or

( ∆σ θ + σ θ )( n 2 + 1)n 2 a 2 − ( n 2 − 1)σ θ n 2 a 2

= rx2 [( ∆σ θ + σ θ )( n 2 + 1) + ( n 2 − 1)σ θ ]

This gives the new inner radius in the form 2 2

2 2

rx = [ 2 n a∆σσ θ(+n∆2 +σ1θ)(+n2σ+1n) n2 a ] 2 θ

(b) 2

rx = [ 2 ( 4 )( 0.0250.1) σσθθ( 5+)0+.12σσθθ((54))4 ( 0.025) ] 2 = 0.02712 = m 27.12 mm ______________________________________________________________________________________ SOLUTION (8.4) Using Eq. (8.13)

σ θ ,max = pi ba −+ab 280 2

= 7 (b02.−6()0.+6b)2

Solving, b = 0.6308 m. Therefore,

t = b − a = 630.8 − 600 = 30.8 mm

______________________________________________________________________________________ SOLUTION (8.5)

( a ) Equation (8.13) and (8.16) give at r=a:

σ θ 1 = pi bb +−aa ,

Then,

σ θ 2 = −2 po b b−a 2

σ θ1 = σ θ 2 ;

pi bb2 +−aa 2 = 2 po b2b−a 2

pi po

= a 22 +bb2 = a22(+44aa )2 = 1.6 2

( b ) By neglecting the strain ε L in the longitudinal direction, Then, gives But

ε θ 1 = E1 (σ θ 1 + νpi ),

ε θ 2 = E1 (σ θ 2 + νpo )

εθ 1 = εθ 2 σ θ 1 + νp i = σ θ 2 − νp o

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.5(CONT.) σθ 1 pi

Hence,

= ba2 −+ab2 = a4 a+2 −4 aa 2 = 1.66 2

σ θ 1 = 1.66 pi

(b)

We also have σθ 2 po

= b22−ba 2 = 42a( 42 −aa )2 = 2.66 2

σ θ 2 = 2.66 po

(c)

Substituting Eqs. (b) and (c) into (a) and letting ν = 1 3 : pi po

= 1.17

______________________________________________________________________________________ SOLUTION (8.6) Equation (8.14), substituting the given data yields

u = apEi ( ba2 −+ab2 + ν ) 2

6 ( 7×10 ) 0.6 + 0.6308 ( 0.63082 −0.62 + 0.3) = 0.200 ×109 2

= 0.426 mm

______________________________________________________________________________________ SOLUTION (8.7) The maximum normal stress is given by Eq. (8.18). Thus, with a = 1 m

σu

a 2 + b2 pi 2 = (σ = θ ) max ns b − a2 Substituting the given numerical values

Hence

280(106 ) 1 + b2 , = (100 ×106 ) 2 2 b −1

b = 2.45 m

t = b − a = 2.45 − 1 = 1.45 m

______________________________________________________________________________________ SOLUTION (8.8) The maximum radial displacement umax occurs at the inner edge of the tank. So, Eq. (8.14) for r=a, results in

umax =

api a 2 + b 2 +ν ) ( E b2 − a 2

Introducing the given data

= umax

(15 ×106 )(0.4) 0.42 + 0.82 −3 ( 2 = = ) m 0.17 mm + 0.3) 0.17(10 9 2 70(10 ) 0.8 − 0.4

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.9) 2

σθ ,max = pi bb +−aa = 53 pi = σ1 , σ r ,max = − pi = σ 2 ( a= ) σu

5 3

pi , = pi

and = σu (b)

σ1 σu

= (340) 204 MPa (governs)

3 5

p= pi 340 MPa i ,

σ2 σ u'

5 pi 3(340)

= − −640pi 1

= − 1;

pi = 154.7 MPa

______________________________________________________________________________________ SOLUTION (8.10) ( a ) We have ε θ = u r , where u is defined by Eq. (8.14). Thus, at r=a:

ε θ ,max = pE [ bb +−aa + ν ] i

(a)

( b ) Introducing σ θ , σ r , and σ z from Eqs. (8.12), (8.13) and (8.20) into Hooke’s law we have

ε θ = E1 [σ θ − ν (σ r + σ z )] 2

ε θ ,max = pE [ b +b(1−−aν ) a + ν ] i

(b)

Substituting the data, Eq. (a): 6

60 (10 ) 4 + a 0.001 = 200 [ + 13 ] (109 ) 4 − a 2 2

Solving, a = 1.41 m. Then,

t =− 2 1.41 = 0.59 m = treq. Similarly, Eq. (b) yields 6

60 (10 ) 4 + 2 a 3 0.001 = 200 [ 4−a 2 + 13 ] (109 )

or = a 1.48 = m; t 0.52 m ______________________________________________________________________________________ SOLUTION (8.11) Equation (8.13) at r=a:

= σ θ ,max

0.252 + 0.052 0.252 − 0.052

= (60) 65 MPa

Equation (8.12) at r=a:

σ r ,max = − pi = −60 MPa

Equation (8.10):

= τ max

60(0.25) = 62.5 MPa

0.252 − 0.052

Equation (8.20) with po = 0 :

= σz

0.05(60) = 2.5 MPa

0.252 − 0.052

Stress-strain relationship is given by

ε θ = E1 [σ θ − ν (σ r + σ z )] = ur

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.11 (CONT.) Substituting Eqs. (8.12), (8.13), and (8.20), this expression results in at r=a:

u = E ( bap2 −i a 2 ) [(1 − 2ν )a 2 + (1 + ν )b 2 ]

(P8.11)

Introducing the given numerical values, we obtain 6

( 60×10 ) [0.4(0.05) 2 + 1.3(0.25) 2 ] u = 72×100.05 6 ( 0.252 −0.052 )

= 0.0571 mm The change in the internal diameter is therefore,

∆d = 2u = 0.1142 mm

______________________________________________________________________________________ SOLUTION (8.12) Equation (8.19): 2

(60) σ θ ,max = − 2(0.25) = −125 MPa 0.25 − 0.05 2

Equation (8.15) at r=b:

σ r ,max = − po = −60 MPa

Equation (8.20) for pi = 0 : 2

0.25 (60) − 0.25 = −62.5 MPa σz = − 0.05 2

Equation (8.9) at r=a and pi = 0 : 2

60(0.25) τ max = − bp −ba = − 0.25 = −62.5 MPa − 0.05 2

Substitution of Eqs. (8.15), (8.16), and (8.20), into Eq. P8.11 of Solution of Prob. 8.11 leads to for r=a: 2

b u = − E (apb2o−a 2 (2 − ν ) ) 6

(P8.12)

×10 )(0.25 ) = − 720.05(60 (2 − 0.3) = −0.0738 mm ×109 (0.252 − 0.052 )

Thus, ∆d =2u =−0.1476 mm ______________________________________________________________________________________ SOLUTION (8.13) Given: b 0.1 = = m c 0.3 = m a 0 δ 0.001(0.1) = = 0.0001 m Using Eq. (8.23): 9

p = Ebδ c 2−cb2 = 200×100.(10.0001) 02.09( 0−.090.01 ) 2

= 88.89 MPa Then, letting = p p= 88.89 MPa, Eq. (8.8) gives at r=b: i

σ θ= 0=

b 2 p − c 2 po 2

c −b

( p − po ) c 2 c 2 −b 2

After substituting the numerical values, this equation results in

po = 49.38 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.14) ( a ) Using Eq. (8.18), σ θ , max pi

= ba2 −+ab2 = 43

b = 2.65a b a +t Then, = = 2.65; a a or

Hence, = t 2a

= t 1.65a

= 0.825

1.65 a 2a

( b ) Neglect longitudinal strain and consider

σr = − pi = 6.3 MPa 4(6.3) = σθ = 8.4 MPa 3

Therefore, for a = 0.075 m and b = 2.65a : 6

×10 )( 0.075 ) 4 ∆d = ε d ( 2a ) = 2 Epi a [ ba2 −+ab2 + ν ] = 2 ( 6.3210 [ 3 + 13 ] (109 ) 2

= 7.5(10−3 ) mm

Alternatively, use Eq. (8.14) and let ∆d = 2u. ______________________________________________________________________________________ SOLUTION (8.15) Introducing various values of P, as given in Fig. 8.4, into Eqs. (a) and (8.21) of Sec. 8.3, it is seen that σ θ and S values as shown in the figure are found. For example, let P = 1 or pi = po . Then,

σ θ = pi 1R− R−1 + pi b 2 ( R1−−11) r ;

And

S = σσθθoi = 11++00 = 1

σ θi = σ θo

σ θ = − pi

S =1

______________________________________________________________________________________ SOLUTION (8.16) ( a ) For ε z = 0; [σ z − ν (σ r + σ θ ) E = 0, using Eqs. (8.8), 2

σ z = ν (σ r + σ θ ) = 2ν ( ab p−−a b p ) 2

( b ) Similarly, for σ z = 0 : 2

ε z = − νE (σ r + σ θ ) = − 2ν (Ea( bp −−ab )p ) i 2

______________________________________________________________________________________ SOLUTION (8.17) At r=a: from Eq. (8.18),

σ θ ,max = 44 aa +−aa pi = 53 pi 2

and from Eq. (8.12),

σ r ,max = − pi .

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.17 (CONT.) Energy of distortion theory: 1

pi [( 53 ) 2 − ( 53 )( −1) + ( −1) 2 ] 2 = σ yp pi = 0.429σ yp

Maximum shearing stress theory: 5 3

pi − ( − pi ) = σ yp

pi = 0.375σ yp

______________________________________________________________________________________ SOLUTION (8.18) We have, at r=a:

σ θ ,max = bb +−aa pi = 99 aa +−aa pi = 45 pi 2

σ r ,max = − pi σ u = 45 pi

(a) or

= pi 0.8(350) = 280 MPa

= σu And

p= pi 350 MPa i ,

( b ) Using Eq. (4.12a), 5 pi 4 ( 350 )

Solving,

− −630pi = 1

pi = 193.8 MPa

______________________________________________________________________________________ SOLUTION (8.19) We have a=20 mm, b=30 mm, c=50 mm. Equation (8.22):

0.03 =

30 p

p 30 + 20 + 30 [ 50 p 29.4 MPa + 0.3] + 10530×10 − 0.3], = 9 [ 302 − 202 502 −302 2

200×109

Steel, Eq.(8.18): c +b + 30 29.4 50 62.5 MPa σ= p= = θ ,max 50 −30 c −b 2

Bronze, Eq.(8.19):

σ θ ,max = −2 p b b− a = −2(29.4) 30 30− 20 = −105.8 MPa 2

______________________________________________________________________________________ SOLUTION (8.20) ( a ) From Eq. (8.18), we have

(0.1) − (0.05) c −b p σ= 25 = 15 MPa = θ ,max b 2 + c 2 (0.05)2 + (0.1)2 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.20 (CONT.) ( b ) Using Eq. (8.35a):

= π bpfl 2π (0.05)(15 ×106 )(0.15)(0.18) F 2= = 127.2 kN

( c ) By Eq. (8.35b):

= T Fb = 127.2(0.05) = 6.36 kN ⋅ m

______________________________________________________________________________________ SOLUTION (8.21) From Eq. (8.18) for r=a: 2

(0.25) − (0.05) = = pi 35 32.31 MPa (0.25)2 + (0.05)2

Total radial force on the contact surface is 2πaLp and total frictional force equals 2πapLµ . Hence, Torque = 2πapLµ ( a )

= 2π (0.053 )(32.31 × 10 6 )(0.2) = 5.075 kN ⋅ m ______________________________________________________________________________________ SOLUTION (8.22) Let ε s1 , ε c1 , and ε s 2 , ε c 2 be initial and final compressive and tensile strains in the shaft and in the cylinder, respectively. Also, let σ θ 1 , p1 and σ θ 2 , p2 denote the initial and final maximum stresses

and contact pressures, respectively. Then, Here,

ε s 1 + ε c1 = ε s 2 + ε c 2

(a)

ε s1 = E1 ( p1 − νp1 ) = 32E p1 ε c1 = E1 (σ θ 1 + νp1 ) = pE ( 2 + 13 ) = 37E p1 1

ε s 2 = E1 [( p2 − νp2 ) + νπpa ] L 2

= E1 [ p2 (1 − 13 ) + 13 π ( 04..055 )2 ] = E1 [ 23 p2 + 19093 .86 ]

From the condition of linearity, σθ 2 σθ 1

and

= pp12 ;

σ θ 2 = p2 σpθ = p2 2pp = 2 p2 1

ε c 2 = E1 (σ θ 2 + νp2 ) = E1 ( 2 p2 + νp2 ) = 73pE

Equation (a) is thus, 2 p1 3E

from which Hence,

+ 73pE1 = E1 ( 23 p2 + 19093 .86 ) + 73pE2

p1 = p2 + 636.62

∆p =p2 − p1 =−636.62 kPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.23) From Eq. (8.22),

a +b δ = Ebp ( bc −+bc + ν b ) + bp E ( b −a − ν s ) b

b[ E1b ( cc2 +−bb2 + ν b ) + E1s ( ba2 −+ab2 − ν s )] 2

(P8.23)

Substituting the given data, this gives p = 9.01 MPa Stresses in the steel cylinder, using Eq. (8.16), 6

×10 )(0.08) − 2(9.01 = −24.03 MPa (σ θ ) r = a = (0.08)2 − (0.04)2 2

) + ( 0.04 ) (σ θ ) r =0.08 = −9.01(10 6 ) (( 00..08 08 ) 2 −( 0.04 ) 2

= −15.02 MPa Stresses in the brass cylinder, from Eq. (8.13): 2

) + ( 0.08 ) (σ θ ) r =0.08 = 9.01(106 ) (( 00..14 14 ) 2 −( 0.08 ) 2

= 17.75 MPa 2(9.01×106 )(0.08)2

= (σ θ ) r =0.14

= 8.74 MPa (0.14)2 − (0.08)2

______________________________________________________________________________________ SOLUTION (8.24) ( a ) Using Eq. (P8.23),

( 0.5 2 )(10−3 )    1−0.29  1 0.152 + 0.12  + 0.33  + 0.1 9  2 2 9  − 72 ( 10 ) 0 . 15 0 . 1   200 (10 )  

= 56.5 MPa ( b ) From Eq. (8.14), for r=b:

= u

2(0.1)2 (0.15)(56.5×106 )

E ( b2 − a 2 )

72×109 (0.152 − 0.12 )

2 a bpi =

= 0.1883 mm = δ 2= u 0.3766 mm ______________________________________________________________________________________ SOLUTION (8.25) ( a ) Using Eq. (8.14), we have at r=a: δ0 a

[(1 − ν )a 2 + (1 + ν )b 2 ] = E ( bpa 2 −a 2 )

from which 2

p = a 2 [(1δ−0νE)(ab2 +−(a1+ν) ) b2 ]

(P8.25)

( b ) Substitution of Eq. (P8.25) into Eqs. (8.12) result in

σ r = (1−ν ) aδ +E(1+ν ) b (1 − br ) 2

σ θ = (1−ν ) aδ +E(1+ν ) b (1 + br ) 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.26) Equation (8.14) at r=b:

u = δ20 = bpE ( bc 2 −+bc2 + ν ) = bpE ( b2.25+ 2b.225−bb2 + 13 ) 2

Solving,

p = 5E.86δ 0b

Then, Eq. (8.17) yields at r=b:

u = bpE (1 − ν ) = 3(25δ.860 ) = 0.114δ 0 Hence,

∆d = 2u = 0.23δ 0

______________________________________________________________________________________ SOLUTION (8.27) ( a ) Intial difference in diameter: where,

Thus,

∆ = 2b(ε 1 + ε 2 )

ε 1 =tangential (comp.) strain in the shaft ε 2 =tangential (tens.) strain in the cylinder ∆ = 2Eb [( p − νp ) + (σ θ ,max + νp )]

= 2Eb [(σ θ ,max + p ) = 2Eb ( 2 p + p ) = 6Epb ( b ) Compressive (uniform) strain ε L , due to axial load P, is

ε L = π bP E 2

We now have

ε 1 = E1 ( p1 − νp1 − πνbP ) 2

ε 2 = E1 (σ θ 1,max + νp1 )

(comp.) (tens.)

where σ θ 1,max is the increased tangential stress. Thus,

∆1 = 2b(ε 1 + ε 2 ) = 2Eb (σ θ 1,max + p1 − πνbP2 ) Based on the linearity condition: σ θ 1, max

σ θ , max P

σ θ 1,max = 2 p1

Setting ∆ = ∆ 1 6 pb E

= 2Eb ( 2 p1 + p1 − 3πPb2 )

P = 9πb 2 ( p1 − p )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.28) Radial strain is

ε = ε b,comp + ε s ,tens. = E1b ( p − νp ) + E1s (σ θ ,max + νp )

We also have

ε = (T2 − T1 )(α b − α s ) = ∆T (19.5 − 11.7)10 −6 = 7.8(10 −6 ) ∆T

Note that from Eq. (8.18) at r=a: σ θ , max p

= 44bb2 +−bb2 = 53

7.8(10 −6 ) ∆T = E1b ( p − 3p ) + E1s ( 53p + 3p )

Thus, or

p = 11( E.7 (+∆3TE) E)10b E6s s

Hence,

E E ( T −T ) σ θ ,max = 1.95 ( E + 3 E )10 2

b s

______________________________________________________________________________________ SOLUTION (8.29) Axial Stress, Eq. (8.18):

= σa

0.82 pi pi a 2 = = 0.8 = pi 80 , b 2 − a 2 1.22 − 0.82

pi = 100 MPa

Tangential stress, Eq. (8.30a):

(σ = t ) max

b2 + a 2 1.22 + 0.82 p pi 2.6 pi 80 , = = = i b2 − a 2 1.22 − 0.82

pi = 30.8 MPa

Shear Stress, Eq. (8.10):

= τ max Solving

pi b 2 1.22 = = = pi 1.8 pi 50 b 2 − a 2 1.22 − 0.82

= pi 27.8 = MPa pall

______________________________________________________________________________________ SOLUTION (8.30) The maximum radial displacement umax occurs at the inner edge of the tank. So, Eq. (8.14) for r=a, results in

umax =

api a 2 + b2 = ( 2 +ν ) E b − a2

Introducing the given data

0.5(60 ×106 ) 0.52 + 0.82 −3 ( 2= = umax + 0.3) 0.456(10 = ) m 0.46 mm 9 2 170(10 ) 0.8 − 0.5

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.31) Equation (8.18) and Eq.(8.12) at r=a:

σ 1 = σ θ ,max = pi ba −+ab = 45 pi

(a) = σ1 − σ 2 (b)

σ yp ;

5 4

σ 2 = σ r ,max = − pi

pi = − (− pi ) 350, = pi 155.6 MPa

σ 12 − σ 1σ 2 + σ 22 = σ yp2 1

pi [( 54 ) 2 − ( 54 )(−1) = + (−1) 2 ] 2 350, = pi 179.3 MPa

______________________________________________________________________________________ SOLUTION (8.32) 2

σθ ,max = pi bb +−aa = 53 pi = σ1 , σ r ,max = − pi = σ 2 ( a= ) σu

5 3

pi , = pi

= σu and (b)

(governs)

p= pi 320 MPa i ,

σ1 σu

= − σσ12' 1;

= (320) 192 MPa

3 5

5 pi 3(320)

= − −620pi 1

pi = 146.6 MPa

______________________________________________________________________________________ SOLUTION (8.33) We have a=20 mm, b=30 mm, c=60 mm. Equation (8.22): 30 p

p 30 + 20 [ 60 +30 + 0.3] + 10530×10 − 0.3], = p 53.3 MPa 9 [ 302 − 202 210×10 602 −302

= 0.05

Steel, Eq.(8.18):

c +b + 30 σ= p= 53.3 60 88.8 MPa = θ ,max c −b 60 −30 2

Bronze, Eq.(8.19):

σ θ ,max = −2 p b b− a = −2(53.3) 30 30− 20 = −191.9 MPa 2

______________________________________________________________________________________ SOLUTION (8.34) ( a ) Use Eq.(8.18) with pi = p,

a = b, b = c:

c +b σ= p= 80 MPa, θ ,max c −b 2

By Eq.(8.22), with = a

= 0.06 or

50 p 100×109 6

c +b = c 2 −b 2

(Fig. 8.6) 6

80(10 ) p

(a)

0,= b 50, = δ 0.06 mm :

p [ 80×p10 + 0.3] + 20050×10 9 [1 − 0.3] 6

60 ×10 =40 ×106 + 0.15 p + 0.175 p,

p =61.5 MPa

( b ) Equation (a) becomes 80 = 61.5

c 2 + 0.052 c 2 − 0.052

= , c 138 mm = 2c 276 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.35) Equation (8.22) with a=0, b=50 mm, c=140 mm. bp

δ = E [ bc −+bc + ν c ] + E [1 − ν s ] c

0.04 =

50 p 120×109

[

1402 + 502 1402 −502

Solving p = 49.4 MPa

p + 0.25] + 21050×10 9 (0.7)

Shaft: σ θ = σ r =− p =−49.4 MPa Cylinder: + 50 c +b = σ= p= 49.4[ 140 ] 63.8 MPa θ ,max c −b 140 −50 2

σ r ,max =− p =−49.4 MPa ______________________________________________________________________________________ SOLUTION (8.36)

δ o from Equation (8.22): bp E

= (ud ) r =b=

[ bc2 −+bc2 +ν = ] 2

bp E

= 1.433 bpE ( 17 15 + 0.3)

Eλ p = 2.866 b

Then, δ i from Eq.(8.22) with a=0,

us =

Therefore,

bp E

(1 −ν ) =

0.7 λ 2.866

= 0.244λ

0.488λ ∆d s =

______________________________________________________________________________________ SOLUTION (8.37) We have (Fig. 8.6):

= a 0.05 = m b 0.1 m c 0.15 m = Maximum tangential stress occurs at r=0.1 m in gear wheel:

σ max = p cc +−bb 2

where, p is the internal pressure exerted by the contact surfaces. Substituting the given data into this equation, we have + 0.1 0.21 = p 00..15 152 −0.12 2

Solving,

= p 0.081 = MPa 81 kPa This interface pressure produces the torque at the contact surface. Area of contact is

2πbL = 2π (0.1)(0.1) = 0.02π

The torque transmitted is thus,

T = [(0.2)(81 × 10 3 )(0.02π )](0.1) = 101.79 N ⋅ m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (8.38) From the torsion, formula, τ = Tb (π b πτb

= T

π (120) b

2) :

= 188.5b3

= 2

Also, Eq.(8.35b):

T = 2πb 2 fpl = 2πb 2 fp(3b) = 6πfb 3 p

And

6π = fb3 p 188.5b3 , = p

= 50 MPa

188.5 6π (0.2)

ν h = ν s = ν , and a=0. Equation (8.22) becomes

We have E h = E s = E , 2

δ = E (2cbpc−b ) = E2(bp4b( 4−bb ) ) = 83pbE 2

8(50) b

Hence = δ

= 0.6667(10−3 )b

3(200×103 )

______________________________________________________________________________________ SOLUTION (8.39) ( a ) a=0, b=10 mm, c=40 mm

Fθ=

T b

= 12 kN

120 0.01

3 12(10 = ) 2= π bfpt 2π (0.01)(0.16)(0.04) p p = 29.84 MPa

Thus or

(Eq.8.24a)

Equation (8.22) gives then

δ= (b)

10(29.84) 100(103 )

+10 + 0.3] + 10(29.84) [ 50 (1 − 0.3)= (4.128 + 1.044)10−3= 0.005 mm 502 −102 200×103 2

+10 c +b σ= = p= 29.84[ 50 ] 32.3 MPa θ ,max c −b 50 −10 2

______________________________________________________________________________________ SOLUTION (8.40) The material properties are (Table D.1)

= = ρ 7860 kN m3 , ν 0.3 GPa, = E 200 = τ yp 210 MPa We have

10,000×2π 2 = ρω 2 7860( = ) 8.62(109 ) 60

The tangential stress, expressed by Equation 8.31b with p=0, has the form

σ= θ

3+ν 8

(a 2 + b 2 − 13++3νν r 2 + arb2 ) ρω 2 2 2

The stresses in the inner and outer edges of the blade are, from the preceding equation,

(σ θ ) r = a=

3.3 8

(σ θ ) r =b=

3.3 8

×20 + 1202 )(10−6 )(8.62 ×109 = (202 + 1202 − 1.93.3 ) 103 MPa is 2

(202 + 1202 − 1.9×3.3120 + 202 )(10−6 )(8.62 ×109 = ) 24.57 MPa The maximum shear stress occurs at the inner surface (r=20 mm), where σ r = 0 : σθ

τ max=

= 51.5 MPa

103 2

The factor of safety, based on the maximum shear stress theory, is then

= n

τ yp τ max

= 4.1

210 51.5

______________________________________________________________________________________ 240 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (8.41) The largest radial stress in the disk, from Equation 8.29, is given by

σ= r ,max

3+ν 8

3.3 8

(b − a ) 2 ρω 2

(120 − 20) 2 (10−6 )(8.62 ×109 = ) 35.6 MPa

The radial displacement of disk is expressed by Equation 8.31c with p=0. Hence,

(= u ) r =b

(3.3)(0.7) 8×200×109

×20 ×20 (202 + 1202 − 1.33.3 )(10−6 )(8.62 ×109 )(0.12) + 1.30.7 2

= 2.3 ×10−5 m = 0.023 mm is the radial displacement at the outer periphery. ______________________________________________________________________________________ SOLUTION (8.42) From Eq. (8.28a), ∂σ r ∂r

= 3+8ν ρω 2 [b 2 + a 2 − ∂∂r ( r 2 ) − a 2 b 2 ∂∂r ( r12 )] = 0

r = ab − 2r = 0; This value of r is substituted into Eq. (8.28a) to yield σ r ,max = ρω 2 (b − a ) 2 2 a 2b2 r3

We also find from Eq. (8.28b), for r=a:

σ θ ,max = 2 3+8ν ρω 2 (b 2 + 13−+νν )

Thus, σ θ , max σ r , max

1−ν 2 a ) 3+ν ( b−a )2

2 ( b2 +

______________________________________________________________________________________ SOLUTION (8.43) At r=0: σ θ = σ r = (3 + ν )b 2

ρω 2 8 . Thus,

2 σ θ − σ rσ θ + σ r2 = σ yp

ω all = b1

8σ yp ( 3+ν ) ρ

(P8.36) 6

×10 ) 2 1 [ 108((2260 ] = 3845.9 rad sec = 0.125 .7×103 ) 3

= 36, 726 rpm ______________________________________________________________________________________ SOLUTION (8.44) We have at inner edge:

τ max = σ θ2−0

= σ θ 2= σ θ 2(90) = 180 MPa Equation (8.28b) with r = 0.03 m is thus 180(106 ) = 108 3 [(0.03) 2 + (0.1) − 101+13 (0.03) 2 + (0.1) 2 ](7800)ω 2 or

from which

rad s 5021 rpm = ω 525.8 = ______________________________________________________________________________________ 241 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (8.45) ( a ) When p = 0, at interface,

ud − u s = 0.05(10 −3 )

(a)

Use Eq. (8.28b) with r=a: 2

(σ θ ) d = ρω4 [(3 + ν )b 2 + (1 − ν )a 2 ] =

7.8(103 )ω 2 4

2 [3.3(0.1252 ) + 0.7(0.025 = )] 101.4ω 2

(σ θ ) d a E

101.4ω 2 a 200(109 )

Radial displacement of disk (with σ r = 0 ):

(ud= )r =a

For shaft, using Eq. (8.30b) with r=a: 2

(σ θ ) s = 3.3( 08.025) (1 − 13..93 )7.8(103 )ω 2 = 0.8531ω 2 Radial displacement of shaft (with σ r = 0 ): 2

ω ( 0.025 ) (u s ) r =a = (σ θE) s a = 0.8531 200 (106 )

Thus, Eq. (a) gives 101.4ω 2 (0.025) 6

200(10 )

ω (0.025) − 0.8531 = 0.05 200(106 )

or = ω 1,994.552 = rad sec 19, 047 rpm ( b ) Maximum stress occurs in disk

2 (σ θ ) max 101.4(1,994.552) 403 MPa = =

______________________________________________________________________________________ SOLUTION (8.46)

a = 0.5c

We have

b = 2c

r = ab = c

( a ) Equation (8.28a): 6 50(10 = ) 3.38c (0.25 + 4 − 1 − 1)7.8(103 )(5000 260π ) 2 or c = 0.1587 m Thus = tr 1.5 = c 238.1 mm 2

( b ) Equation (8.28b) at r=a:

σ θ ,max = 3.38c [0.25 + 4 − 13..93 (0.25) + 4]7.8(10 3 )(5000 260π ) 2 2

= 180.1 MPa ______________________________________________________________________________________ SOLUTION (8.47) ( a ) Equation (8.23), with a=0:

= p

210(109 )(0.000075) 0.16 − 0.005625 2(0.075) 0.16

= 101.31 MPa

Then, Eq. (8.18) gives

6 0.16 + 0.005625 = σ θ ,max 101.31(10 = ) 0.16−0.005625 108.69 MPa

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.47 (CONT.) ( b ) Applying Eq. (8.28c),

.3( 0.7 ) 0.000075 = 8(3210 [0.005625 + 0.16 − 13..33 (0.005625) ×109 )

+ 10..37 (0.16)](7.8 × 10 3 )(0.075)ω 2

Solving,

= = ω 450 rad sec 4300 rpm ______________________________________________________________________________________ SOLUTION (8.48) From Eqs. (8.30a) and (8.30b), for r=0, υ = bω , and ν = 1 3 :

σ max =

ρ (ωb) 2 =

3+ν 8

5 12

ρυ 2

______________________________________________________________________________________ SOLUTION (8.49) Apply Eq. (8.28c) with u = 0.04 mm at r = 25 mm . Substituting the given data:

.3( 0.7 ) 0.04(10 −3 ) = 8(3210 [(0.025) 2 + (0.25) 2 − 13..33 (0.025) 2 ×109 ) 2 2 + 1.3 0.7 (0.25) ](7800)ω (0.025)

Solving,

= = ω 913.1 rad s 8719 rpm ______________________________________________________________________________________ SOLUTION (8.50) We have ti to

= ( ab ) s ;

0.125 0.0625

s = ( 00..625 125 )

Solving, s=0.431. Then, 1

0.431 2 2 m1, 2 = − 0.431 2 ± [( 2 ) + (1 + 0.3 × 0.431)]

m1 = 0.869

m 2 = − 1.3

Equation (8.39) gives

σ r = ct r 0.3 + ct r −1.869 − 0.50169 ρ (ωr ) 2 1

Given conditions are:

(σ r ) r =0.625 = 0,

(a)

(σ r ) r =0.125 = 0

Thus, substituting the given data into Eq. (a) and solving: c1 t1

= 0.2322 ρω 2 ,

c2 t1

= −0.0024 ρω 2

The second of Eqs. (8.39), at the bore r=0.125 m, leads to

140(10 6 ) = 0.2322(0.1250.3 )(0.869) ρω 2

) 2 + ( −0.0024)( −1.3)(0.125−1.869 ) ρω 2 − (81−+(03.9.3)()(00.125 .431) ρω

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.50 (CONT.) or

ρω 2 = 548(106 )

It follows, from the second of Eqs. (8.39), that

(σ θ ) r =0.625 = [0.2322(0.6250.3 )(0.869) 2

6 −0.0024(−1.3)(0.625−1.869 ) − 81.9(0.625) −3.3(0.431) ](548 × 10 )

= 38.3 MPa Circumferential force is thus

38,300(0.0625) = 2394 kN m

______________________________________________________________________________________ SOLUTION (8.51) ( a ) Uniform thickness

Centrifugal force due to the blades, mω r , is 2

540 10,000π 2 9.81 60

) (0.575) = 8.677(106 ) N

(

The pressure at b is then

= po We have

8.667(106 ) 2π (0.5)(0.05)

= 55.18 MPa

000π 2 ρω 2 = 7.8(10 −3 )( 10,60 ) = 2.1384(10 3 )

The condition of zero pressure at the bore is satisfied by, using Eq. (8.27b): 2

2 2

(σ r ) r =a = 0 = 1−Eν 2 [ − ( 3+ν )(1−8νE ) ρω a + (1 + ν )c1 − (1 − ν ) ac22 ] or

0 = − ( 3+ν )8a ρω + [ E1(−1ν+ν2 ) c1 ] + [ − E1(−1ν−ν2 ) c2 ] a12     A1

0 = A1 + A2 a12 −

( 3+ν ) a 2 ρω 2 8

Substituting the data, we have

0 = A1 + 256 A2 − 3.4456

(a )

The condition at the outer circumference is satisfied by: 2

A2 3.3(0.5) (2138.4) (σ r ) r =b = 55.18 =A1 + (0.5) 2 − 8

55.18 =A1 + 4 A2 − 220.52

(b)

A1 = 280.

(c)

Solution of Eqs. (a) and (b) gives

A2 = −1.08

Equation (8.27c) is also written as follows 2 2

σ θ = A1 − A2 r1 − (1+3ν )8ρω r

(d)

where A1 and A2 are given by (c). At r=a, we have from Eq. (d):

(σ θ ) r = a = 280 + 276.48 − 1.98

= 554.5 = MPa σ max

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.51 (CONT.) Similarly, Eq. (d) gives r=b:

(σ θ ) r =b = 280.08 + 4.32 − 126.97 = 157.35 MPa

at r Note that σ r ,max occurs =

= ab 0.1768 m. Thus Eq. (8.27):

(σ r ) r =0.1768 = −27.67 + 280 − 34.55 = 217.78 MPa ( b ) Hyperbolic section We have Then,

0.4 0.05

.5 = ( 0.00625 )s ;

s =1 1

m1, 2 = − 12 ± [0.25 + 1.3] 2 or Letting

m1 = 0.745 c1 t1

m2 = −1.745 c2 t1

= B1

= B2

Eq. (8.39) becomes then 2

2 σ r = B1r m + s −1 + B2 r m + s −1 − (83−+(ν3)+ρω ν )s r 1

(σ r ) r =a = 0

and or Also or

= B1 (0.0625) 0.745 + B2 (0.0625) −1.745 − 0.7021(0.0625) 2 ρω 2 (e) 0 = B1 + 996.35B2 − 0.0216 ρω 2 (σ r ) r =b = 55.24 − B1 (0.5) 0.745 + B2 (0.5) −1.745 − 0.702(0.5) 2 ρω 2

55.24 = 0.55967 B1 + 3.352 B2 − 0.1755ρω 2

(f)

Substituting the given data and solving Eqs. (e) and (f):

B1 = 725.14

B2 = −0.6814

(g)

The second of Eqs. (8.33),

σ θ = B1m1r m + s −1 + B2 m2 r m + s −1 − (18+−3(ν3+)νρω) s r 2 1

gives at r=a:

(σ θ ) r =a = 725.14(0.745)(0.0625) 0.745

− 0.6814( −1.745)(0.0625) −1.745 − 0.4043(0.0625) 2 ρω 2 or

= = (σ θ ) r = a 215.2 MPa σ max

Also similarly we obtain that

(σ θ ) r =b = 110.2 MPa

(CONT.) ______________________________________________________________________________________ 245 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 8.51 (CONT.) ( c ) Uniform stress Using Eq. (8.41), where,

to t1

= e − ( ρω 2σ ) b

−7.8×10 (523.59) (0.5) −( ρω 2 2σ )b 2 = 2(84×106 )

= −3.182 Thus, to 0.02425 e 0.001 m 1 mm = = = −3.182

−12.728 r 2

and t = 0.02425e ______________________________________________________________________________________ SOLUTION (8.52) Because u=0 at r=b, set c2 = 0 in Eq. (8.43). Thus, Eq. (8.43) with c2 = 0, T =constant, and u=0 at r=b: b

0 = c1b + (1+bν ) αT ∫ rdr

c1 = − (1+ν2)αT

Substituting this into Eq. (d) and (e) of Sec. 8.10, we obtain 2 αET σ r = − αET 2 − 2 (1−ν ) (1 + ν ) 2

2 αET σ θ = αET 2 − αET − 2 (1−ν ) (1 + ν ) 2

Letting 1 − ν

= (1 + ν )(1 − ν ) and simplifying we obtain σ r = σ θ = − α1ET −ν 2

______________________________________________________________________________________ SOLUTION (8.53) We substitute the given T into Eqs. (8.44) to obtain

= σr Then,

∂σ r ∂r

α E (Ta −Tb ) 2ln( b a )

[− ln ba − ra2 ((br2 −−ab2 )) ln ba ]

Eα Ta −Tb d b a d b b 0= = 2ln( b a ) { dr (ln a ) − b 2 − a 2 [1 − dr ( r 2 )]ln a } (

)

yields, after differentiation:

r = ab( b2 −2a 2 ln ab ) 2

(P8.53)

It is noted that, Eq. (8.53) gives the same result. ______________________________________________________________________________________ SOLUTION (8.54) Using Eq. P8.53 of Solution (8.53), we have 1

2ln(1.5) = r 0.01(0.015)[ = ] 2 12.1 mm (0.015)2 − (0.01)2

Equation (8.53) are therefore −6

(12.1 −15 ) (σ r ) r =12.1 = 10.4 (102 ( 0.7)()90ln(×110.5))( −8) [ − ln 1215.1 − 100 ln(1.5)] 12.12 (152 −102 )

= −1.319(107 )[−0.041] = 0.541 MPa (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 8.54 (CONT.) 2

−15 ) −1.319(107 )[− ln(1.5) − 100(10 (σ r ) r =10 = ln(1.5)] 100(152 −102 )

= 0 MPa 2

+15 ) −1.319(107 )[1 − ln(1) − 100(15 (σ θ ) r =15 = ln(1.5)] 152 (152 −102 )

= −4.64 MPa 2

+15 ) 15 −1.319(107 )[1 − ln( 12.1 (σ θ ) r =12.1 = ) − 100(12.1 ln(1.5)] 12.12 (152 −102 )

= 0.497 MPa Similarly, 2

(σ z ) r =10 = −1.319(10 7 )[1 − 2 ln(1.5) − 152 (210−10)2 ln(1.5)]

= 6.063 = MPa σ max

(σ z ) r =15 = −1.319(107 )[1 − 2 ln(1) − 152 (210−10)2 ln(1.5)]

= −4.63 MPa

______________________________________________________________________________________ SOLUTION (8.55) Introducing

T ( r ) = T0 (b − r ) b

into Eqs. (8.45), after integration, we obtain the following expressions for stresses

σ r = 13 T0 ( br − 1)αE σ θ = 13 T0 ( 2br − 1)αE

From these we observe that, at r=b: the radial stress vanishes while tangential stress assumes its maximum value. ______________________________________________________________________________________ SOLUTION (8.56 through 8.59) Owing to the symmetry, only any one-quarter of the cylinder need be analyzed. Use a finite element computer program, such as ANSYS.

End of Chapter 8

CHAPTER 9 SOLUTION (9.1) Using Eq. (9.3), 1 4

β = ( 4 kEI ) = [ 4 ( 200×101.4 ()(105.04) ×10 ) ] 9

−6

1 4

= 0.7676 m −1 Equation (9.8) yields, for x = 0 :

M max = −4 βP f 3 ( βx ) = − 4Pβ f 3 (0) = − 4Pβ

Thus, 6

−6

10 ) 4 ( 0.7676 ) P = σ max Ic( 4 β ) = 210×10 ( 5.040.×0635 = 51.18 kN

______________________________________________________________________________________ SOLUTION (9.2) b

I = b ( 2.5b) 3 12 = 1.302b 4 6

2.5b

I (1.302 b ) M max = σ max = 250×101.25 = 260.4(10 6 )b 3 c b 6

(a)

1 4

β = [ 4 ( 200×2010(10)(1.)302 b ) ] = 0.0662 b 9

Equation (9.8):

M max = 4Pβ

(b)

From Eqs. (a) and (b), 3

260.4(10 6 )b 3 = 440( 0(.100662) b) or

= b 0.0241 = m 24.1 mm

______________________________________________________________________________________ SOLUTION (9.3) Through the use of Eq.(9.3), we have 1

k 4 = β (= [ 4(210×1015 = ] 4 0.817 in.−1 The maximum deflection and 3 4 EI ) )(40)10−6 )

moment takes place under the load and hence f= 1 Applying Eqs.(9.8). We obtain:

= vmax

Pβ 2k

= f1

(150)(0.817) 2(15)

= (1) 4.09 mm

M= max

P 4β

150(10 ) 4(0.817)

σ= max

M max c I

45.9×103 (0.1)

= f3

f= 1.0 . 3

= (1) 45.9 kN ⋅ m

= 114.8 MPa

40(10−6 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.4) Moment of inertia of the beam equals 90(180)3 12

= 43.74(106 ) mm 4

bh3 12

= I

From Eq.(9.3), we obtain

5(10 ) = β [= ] 4 1.242 m −1 4(12×109 )(43.74×10−6 )

Apply Eq.(c) of Example 9.1, with f 4 ( β a ) = p 2k 40 2(5000)

f 4 ( β b) = −0.0655 from Table 9.1:

vmax = [2 − f 4 ( β a ) − f 4 ( β b)] =

[2 − 2(−0.0655)] = 8.524 mm

______________________________________________________________________________________ SOLUTION (9.5) Select a particular solution of the form v p = a sin 2Lπx a = const. Introduce this into Eq. (9.1),

( 2Lπ ) 4 a + EIk a = EIp1

or Thus,

= k [1+ 4 (pπ1 βL )4 ]

16π 4 EI L4

p1 16π 4 EI

sin 2Lπx

(a)

General solution is

v = e βx [ A cos βx + B sin βx ] + e − βx [c cos βx + D sin βx ] + v p

Boundary conditions are:

v ( ∞) = 0;

A= B=0 Loading repeats itself periodically, p1 sin( 2πx L). But

v = e − βx [C cos βx + D sin βx ] cannot repeat periodically and represents a damped wave. Thus, in order deflection to repeat itself with the same wave length as loading it is required that C = D = 0. Accordingly, solution is

v = vc + v p = 0 + v p = v p

given by Eq. (a). ______________________________________________________________________________________ SOLUTION (9.6) p Q x a b

From Example 9.1: b

− βx − βx vQ = ∫ pdx [cos βx + sin βx ] − ∫ pdx [cos βx + sin βx ] 2 k βe 2 k βe

0 p 2k

− βx

cos βa − e

− βb

cos βx ]

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 9.6 (CONT.) Note that, if a=0 and b=L (large):

vQ ≈ 2pk

When a and b increase (large):

vQ → 0

(a)

While according to the result of Example 9.1, when a=0 and b=L (large):

vQ ≈ 2pk

When a and b increase (large);

vQ ≈ 2pk

(b)

the answers differ, as observed by comparing Eqs. (a) and (b). ______________________________________________________________________________________ SOLUTION (9.7) Applying Eqs. (9.3) and (9.8) we obtain 1

16.8 k 4 = β [= [ 4(8.437) = ] 4 0.84 m −1 4 EI ]

v (0) = v max = P2βk =

0.135(0.84) 2(16.8)

= 3.375(10−3 ) m

M (0) = M max=

and

P 4β

= 40.179 kN ⋅ m

135 4(0.84)

Maximum stress is therefore M max S

σ= max

40.179(103 )

= 103.02 MPa

3.9(10−4 )

______________________________________________________________________________________ SOLUTION (9.8) Refer to Solution of Prob. 9.1:

1 k 14 12(106 ) 4 ) [ ] = β (= 4 EI 4(200 ×109 )(5.04 ×10−6 ) = 1.3135 m −1 Equation (9.8) gives, for x = 0 : P M max = − 4β

Therefore

P =

σ max I (4 β )

= c

(250 2.5)106 (5.04 ×10−6 )4(1.3135) = 41.7 kN 0.0635

______________________________________________________________________________________ SOLUTION (9.9)

I = b 4 12

M = max

σ max I

= c

(260 1.8)106 (b 4 12) b2

(a)

= 24.074(106 )b3 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 9.9 (CONT.) 1 7(106 ) ] 4 0.1316 b = β [= 9 4 4(70 ×10 )(b 12)

Equation (9.8):

M max =

P 4β

(b)

From Eqs. (a) and (b),

50(10)3 b 4(0.1316) Solving, = b 0.0628 = m 62.8 mm 24.074(106 )b3 =

______________________________________________________________________________________ SOLUTION (9.10) We have

1 k 14 15 ] [ ] 4 0.8165 m −1 = = β [= 4 EI 4(8.437) Pβ v= (0) v= max 2k 0.25(0.8165) −3 ) m 6.8 mm = = 6.8(10 = 2(15) 250 P And M (0) = M max = = = 76.55 kN ⋅ m 4 β 4(0.8165)

Maximum stress is thus

= σ max

M max 76.55(103 ) = = 196.3 MPa S 3.9 ×10−4

______________________________________________________________________________________ SOLUTION (9.11) Equations (9.3) and (9.8): 1 4

β = [ 4 kEI ] , (a)

(b)

v max = P2βk ,

M max = 4Pβ

k is 1.25 times the actual value.

β changes by 4 1.25 = 1.057. v max changes by 1.057 1.25 = 0.846. σ max (or M max ) changes by 1 1.057 = 0.946. k is 1.4 times the actual value.

β changes by 4 1.4 = 1.088. v max changes by 1.088 1.4 = 0.777. σ max (or M max ) changes by 1 1.088 = 0.919.

Note that stress calculation is not affected appreciably in both cases. ______________________________________________________________________________________ 251 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ P1 P2 SOLUTION (9.12) O

x 0.83 m

0.83 m

y By using the principle of superposition, deflection of any point, say O, of rail is expressed as the algebraic sum of v1 and v 2 caused by P1 and P2 , respectively. Thus, from Eq. (9.8), we have

v0 = P21kβ f1 ( βx1 ) + P22kβ f1 ( βx 2 )

(a)

. Then, f1 ( βx1 ) and f1 ( βx2 ) are found from Table 9.1 for x1 = 0.83 m and x2 = −0.83 m. We have βx1 = 0.697 and βx2 = −0.697. Thus f1 ( βx1 ) = 0.702. Since, from symmetry f 1 ( βx1 ) has the same value for a (+) or (-) value of βx ,

where, P1 = P2 = P, β = 0.84 m

−1

Eq. (a) may be written as

v0 = P2βk (0.702 + 0.702) = 1.404 P2βk

Resultant bending moment at O, from Eq. (9.8), is

M 0 = 4Pβ f 3 ( βx1 ) + 4Pβ f 3 ( βx 2 )

= 4Pβ [2 f 3 (0.697)] = 0.125 4Pβ It may be verified by comparing the results of Problems 9.7 and 9.9 that addition of one or more load (reduces appreciably value of maximum moment) causes a large increase in the maximum deflection of the rail. ______________________________________________________________________________________ SOLUTION (9.13) x px B

Expression for loading are:

p x = pL0 ( a − x ) Px =

p0 L

(a + x )

Q y

(segment BQ) (segment QC)

Deflection at Q is obtained by substituting ( p x dx ) for P in Eq. (9.6). That is a

vQ = 2p0kLβ {∫ ( a − x )e − βx [cos βx + sin βx ]dx 0

Integrating, we have

= vQ

p0 4 β kL

+ ∫ ( a + x )e βx [cos βx + sin βx ]dx} 0

[ f3 ( β a ) − f3 ( β b) − 2 β Lf 4 ( β b) + 4 β a ]

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.14) p A

x y p

Figure (a)

p/k x

Figure (b)

Figure (c)

A R

Referring to Fig. (b):

θ A1 = 0

v A1 = − kp

Referring to Fig. (c) and using Eq. (9.12),

v A2 = 2kβ [ − Rf 4 (0) + βM 0 f 3 (0)] 2

θ A2 = − 2 βk [− Rf1 (0) + 2 βM 0 f 4 (0)] The problem may be separated into two different cases both semi-infinite beams (as is seen in the figure above) provided that deflection at A is zero. Thus,

solving,

v A1 − v A2 = 0;

− kp = 2kβ [ − R + βM 0 ]

θ A1 − θ A2 = 0;

2β 2 k

M 0 = 2 β 2 EI kp ,

[ − R + 2 βM 0 ]

R = 4 β 2 EI kp = βp

______________________________________________________________________________________ SOLUTION (9.15) ( a ) Using Eq. (9.12), v = 2

v max = 2 β kM A

2β 2 k

M A f 3 ( βx )

at x = 0

(down)

Scan Table 9.1, f 3 ( βx ) = −0.2079 at βx = π2 , and 2

v min = 2 βk M A ( −0.2079)

(up)

Therefore

vmax vmin

1 = −0.2079 = −4.81

( b ) Applying Eq. (9.12), M = M A f 1 ( βx )

v max = M A

at x = 0

Scan Table 9.1, f 1 ( βx ) = −0.0432 at βx = π , and

M min = −0.0432 M A

Thus

M max M min

1 = −0.0432 = −23.15

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.16) ML x

A y Equation (9.11) with v (0) = 0 gives

0 = 2 β ( − R A + βM L ) k from which R A = M L . When R A is substituted for − P into Eq. (9.12): − 2 βM2LEI e − β sin β x = − 2 βM2LEI f 2 ( β x) v= Successive differentiations of this expression give:

v ' = − 2Mβ EIL f3 ( β x) M = 2MEIL f 4 ( βx ) V = − 2MEIL f1 ( βx )

______________________________________________________________________________________ SOLUTION (9.17) Ks a

We have = k

= 288 kPa

180 0.625

1 4

) β = [ 4(200×288(10 ] = 0.51697 m −1 10 )(5.04×10 ) 9

Using Eq. (9.8),

Pβ 2k

v(0) = vmax =

M (0) = M max= Thus,

σ max=

6.75(0.51697) 2(288)

= 6.06 mm

= 3.26421 kN ⋅ m

6.75 4(0.51697)

P 4β

3264.21(0.625)

Mc I

5.4×10−6

= 377.8 MPa

______________________________________________________________________________________ SOLUTION (9.18) We = have k

120 = 1.2 100 kPa . Equation (9.3) is thus 1

k 4 = = β [= [ 4(70×106100 ] 4 0.6148 m −1 4 EI ] )(2.5×10−6 )

Check: Length of spring spacing: π a<= 4β

= 1.277 m

4(0.6148)

OK.

Check: Length of beam:

3π L = 8.5 > 23πβ = 2(0.6148) = 7.665 m

OK.

The largest deflection and moment of the beam take place under the load and hence f= f= 1.0 . 1 3 Applying Eqs.(9.8), we obtain

= vmax

M= max

σ max =

Pβ 2k

15(0.6148) 2(100)

= f1

= f3

P 4β

Pc 4β I

= (1) 0.0461 mm

15(103 ) 4(0.6148)

= (1) 6.1 kN ⋅ m

15(10 )(0.06)

= 146.4 MPa

4(0.6148)(2.5×10−6 )

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.19) From Eq. (9.3), 3

18(10 ) = β [= ] 4 0.824 m −1 4(0.375×200×109 )(5.2×10−7 )

Since

βL = 0.824(0.74) = 0.618 < π 4 the beam can be considered rigid.

( a ) Uniform deflection is

= v

F 3K

= 10 mm

540 3(18)

P K

( b ) We may replace the given beam by the beams shown in Figs. (a) and (b). 0.25

0.125

0.375 Figure (a)

P F+PB1

F+PB2

F+PB3

Figure (b)

Note that F is the direct load on each spring and PBi is the bending loads with the values:

= F P= 3 540= 3 180 N PBi = M Brr2i ∑i

and Here

= PB1

(i = 1, 2, 3)

540(0.125)(0.375)

= 90 N , PB 2 = 0, 2(15×0.125)2

PB 3 = − PB1

End deflections are thus,

vL = v1 = (180 + 90) 18 = 15 mm vR = v2 = (180 − 90) 18 = 5 mm

______________________________________________________________________________________ SOLUTION (9.20) Using Eq. (9.3), 1

14 4 = β [= 0.8034 m −1 4(8.4) ]

= β L 0.8034(0.6) = 0.482 = rad 27.62o Hence, from Eq. (9.13):

( 0.8034 ) 2 + cos 27.62 + cosh 27.62 v c = 4500 2 (14×106 ) sin 27.62o +sinh 27.62o o

−6 = 536(10 = ) m 0.536 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.21) P

βL < 1

E C L/2

L/2

Referring to Sec. 9.8: 4

5[( P + pL ) L ] L 5 pL vc = 48PLEI + 384 EI − 384 EI 3

(0.15)3 48(8.4×106 )

[9000 + 5(75008×0.15) − 5(90008+1125) ]

= 2.83(10−8 ) m Similarly, 3

0.15](0.15) θ E = 16PLEI + 24pLEI − [(9000+1125) 24 EI 2

Substituting the given data,

θ E = 5(10−7 ) rad

______________________________________________________________________________________ SOLUTION (9.22) Equation (9.3),

k 14 8 14 ] [ = ] 0.6687 m −1 = β [= 4 EI 4(10) rad 30.65o = β L 0.6687(0.8) = 0.535 =

From Eq. (9.13):

15(103 )(0.6687) 2 + cos 30.65o + cosh 30.65 2(8 ×106 ) sin 30.65o + sinh 30.65 = 626.9[3.7425](10−6 ) −6 = 2,346(10 = ) m 2.35 mm vc =

______________________________________________________________________________________ SOLUTION (9.23) Refer to Sec. 9.8 and Table D.4, we have = I

3 bh = 12 46.8(10−6 ) m 4 ,

= EI 3.4 MN ⋅ m 2 , and β L < 1 .

PL3 5 pL4 5[( P + pL) L]L4 vc = + − 48 EI 384 EI 384 EI 3 (0.4) 5(7000 × 0.4) 5(8000 + 2800) = [8000 + − ] 6 48(3.4 ×10 ) 8 8 −3 = 0.00118(10 = ) m 1.18(10−3 ) mm (CONT.) ______________________________________________________________________________________ 256 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 9.23 (CONT.) Likewise,

PL2 pL3 [(8000 + 2800) 0.4](0.4)3 + − 16 EI 24 EI 24 EI 2 (0.4) 7000(0.4) 10,800 = [8000 + − ] 6 16(3.4 ×10 ) 32 32

θE =

= 7.84(10−6 ) rad ______________________________________________________________________________________ SOLUTION (9.24) L p x Figure (a) y -v1 -v1 z 0 v1 Figure (b) -v2 -v1 v1=v2 z Figure (c) 0 v1 v2 -v1 -v3 v1=v3 z Figure (d) 0 v1 v2 v3 Boundary conditions are:

v (0) = v ( L) = 0,

v ' ' ( 0) = v ' ' ( L ) = 0

(a)

These are transformed into the following central difference conditions by using Eq. (7.4) and (7.7):

v0 = 0,

− v −1 = v1 ,

For m=2 (Fig. b):

v n = 0,

v n −1 = −vn +1

(b)

vn −2 − 4vn −1 + 6( mm4+1 )vn − 4vn +1 + vn + 2 = 1m.64

v1 − v1 −v1 + 6 224+1= 4

For m=3 (Fig. c):

1.6 24

;= v1 21.4 mm

− 4v1 + 6 334+1 v2 − v2 = 13.46 4

− v1 + 6 334+1 v1 − 4v2 = 13.46 4

Solving, v= 1

v= 17.2 mm 2

For m=4 (Fig. d):

− v1 + 6 444+1 v1 − 4v1 + v1 = 14.46 4

− 4v1 + 6 444+1 v 2 − 4v1 = 14.46 4

Solving, v= 3

v= 10.2 mm 1

v2 = 13.9 mm

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (9.25) From Example 9.1:

v p = 2pk ( 2 − e − βa cos βa − e − βa cos βb)

where

k = 48aLEI3

Thus,

− βa t cos βa − e − βb cos βb) v p = 2 (paL 48 ) EI [ 2 − e

= 96paLEIt [2 − f 4 ( βa ) − f 4 ( βb)] We= have β

= 3.936 6 0.656 m −1

At midspan, we have βa = βb Thus,

v p = 48 EI

Since,

v m = 48paLEIt = R48ccEILt

Solving,

and

f 4 ( βa ) ≈ 0

paL3t

Rcc = ap = 0.3 p End of Chapter 9

CHAPTER 10 SOLUTION (10.1) P

B x

D x

A Defections at point D: We write

M DB = Px

Using Eq. (10.5),

M BA = Pb + Rx

U DB = ∫ [ ( 2PxEI) + 2RAE ]dx = P6 EIb + 2RAEb 2

0 a

2 3

U BA = ∫ [ ( Pb2+EIRx ) + 2PAE ]dx 2

= 2 1EI [ p 2 ab 2 + PRa 2 b + 13 R 2 a 3 ] + 2PAEa 2

Total strain energy U = U BD + U BA . Vertical deflection at D is thus Pa δ v = ∂∂UP = EI1 [ Ra2 b + P( ab 2 + b3 )] + AE 2

Horizontal deflection at D:

Rb δ h = ∂∂UR = EI1 [ Ra2 b + Ra3 ] + AE 2

Angular rotation of D: Introduce a couple moment C at D as shown in the figure. Then, and

M DB = Px + C b

M BA = Pb + Rx + C

U DB = ∫ [ ( Px2+EIC ) + 2RAE ]dx 2

= 2 1EI [ P 3b + C 2 b + PCb 2 ] + 2RAEb 2 3

U BA = 2 1EI ∫ [ Pb + Rx + C ]2 dx + 2RAEb 2

= 2 1EI [ P 2 ab 2 + R 3a + Ca 2 + PRa 2 b + 2 PCab + CRa 2 ] + 2PAEa 2 3

Hence,

θ D = ∂∂UC C =0 = 2 1EI [ Pb 2 + 2 Pab + Ra 2 ]

Note that displacements of a simple (straight) cantilever beam may be readily found by setting b=0 in the foregoing results. ______________________________________________________________________________________ 259 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (10.2) Equations of statics are applied to obtain the reactions:

∑ F =0 : ∑ M =0 : ∑ M =0 :

R Ax = 0

RC = 14 P1 + 12 P2 + 43 P3

R Ay = 43 P1 + 12 P2 + 14 P3

The axial force in each member is obtained by applying the method sections, as required. The results are as follows:

N DC =

2 4

P1 + 22 P2 + 3 4 2 P3

N AE = 3 4 2 P1 + 22 P2 + 42 P3 N DE =

1 2

P1 +

P2 +

1 2

N BE = − 42 P1 + 22 P2 + 42 P3 N BD = N AB = N BC =

P1 + 22 P2 − 42 P3 3 1 1 4 P1 + 2 P2 + 4 P3 3 1 1 4 P1 + 2 P2 + 4 P3

2 4

Numerical results are determined for P= 1

These are tabulated bellow.

Bar Axial force (kN)

N DC = 99.5

AE DE

N AE = 99.5 N DE = 90

BE BD AB BC

N BE = 31.8 N BD = 31.8 N AB = 67.5 N BC = 67.5

P= P= 45 kN and L = 3 m. 2 3

∂N ∂P2

Length (m)

2 2 2 2 1 2 2 2 2 12 12

2.125 2.125 3.000 2.125 2.125 3.000 3.000

Thus, the vertical deflection at B: 7

δ B = AE1 ∑ N j ∂P L j j =1

∂N j 2

= 10AE [2(67.5)(0.5)(3) + 2(31.8)(0.707)( 2.125) + 90(1)(3) + 2(99.5)(0.707)2.125] = 867, 023.675 AE m 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.3) The moment is expressed by

M = P( 2a − x );

∂M ∂P = 2a − x

Castigliano’s theorem gives then or

2 M ∂M P v P = ∫ EI ∂P dx = ∫ E ( 2a − x ) ( c1 x + c2 ) dx + ∫ 4

P EI 2

( 2a − x ) 2 dx

v P = 1112PcE1a + 7 Pc3 E2a + 3PaEI 2 3

______________________________________________________________________________________ SOLUTION (10.4) P C

y Deflection at A

(with M = Px ):

v A = EI1 ∫ M ∂∂MP dx = EI1 ∫ Px ( x )dx 0

vA =

PL3 3 EI

(with M = Px + C ):

Slope at A

θ A = EI1 ∫ M ∂∂MC dx = EI1 ∫ ( Px + C )(1)dx

Setting C = 0 and integrating

θ A = 2PLEI 2

______________________________________________________________________________________ SOLUTION (10.5) p

P C

y Deflection at A

B 2

(with M = Px + 12 px ): L

δ A = EI1 ∫ M ∂∂MP dx = EI1 ∫ ( Px + 12 px 2 )( x )dx 0

= Slope at A

PL3 3 EI

pL4 8 EI

(with M = Px + 12 px

+ C ):

θ A = EI1 ∫ M ∂∂MC dx = EI1 ∫ ( Px + 12 px 2 + C )dx

Setting C = 0 and integrating

θ A = 2PLEI + 6pLEI 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.6) P C

A y

Deflection at A

I B L/2

L/2

(with M = Px ):

δ A = EI1 ∫ M ∂∂MP dx = EI1 ∫ ( Px )( x )dx + 2 1EI ∫ ( Px )( x )dx Integrating,

7 PL 3 PL δ A = 24PLEI + 48 EI = 16 EI 3

(with M = Px + C ):

Slope at A

θ A = EI1 ∫ M ∂∂MC dx = EI1 ∫ ( Px + C )(1)dx + 2 1EI ∫ ( Px + C )dx Setting C = 0 and integrating 3 PL θ A = 8PLEI + 16 EI

5 PL θ A = 16 EI

______________________________________________________________________________________ SOLUTION (10.7)

R ds=Rdθ

P ( a ) Horizontal deflection

M θ = PR (1 − cos θ ) + QR sin θ + M 0

and

δ h = EI1 ∫ M θ ∂∂MQθ ds = EI1 ∫

π 2

R2 2 EI

[ PR 3 (1 − cos θ ) sin θ + QR 3 sin 2 θ + M 0 R 2 sin θ ]dθ

( 12 PR + M 0 )

Vertical deflection (Q=0)

δ v = EI1 ∫ M θ ∂∂MPθ ds = EI1 ∫

π 2

[ PR 3 (1 − cos θ ) 2 + MR 2 (1 − cos θ )]dθ

= 4REI [ PR(3π − 8) + 2 M 0 (π − 2)] 2

( b ) Rotation of the free end (Q=0)

π 2

θ = EI1 ∫ M θ ∂∂MMθ ds = EI1 ∫ [ PR(1 − cos θ ) + M 0 ]dθ 0

= 2REI [ PR(π − 2) + πM 0 ]

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.8) We now have

M = − FR sin θ ,

N = F sin θ , Applying Eq. (10.6), with α = 6 5 and T = 0 :

V = − F cos θ

2 2 R δ h = REI ∫ F sin 2 θdθ + AE ∫ F sin θdθ + 56AGR ∫ F cos θdθ 3

Integrating,

3πFR δ h = π2FREI + π2FR EA + 5 AG 3

Substituting the given data: 3

)( 0.05 ) 4000 )( 0.05 ) )( 0.05 ) δ h = π ( 4000 + π (400 + 3π5((4000 400 ( 6.67 ) (10 ) 2 80×10 ) 2 5

= (0.59 + 0.008 + 0.02)10 −3 = 0.62 mm The error, if N and V are omitted, is 4.5 %. ______________________________________________________________________________________ SOLUTION (10.9) Introducing a rightward horizontal force Q at point D, we write

0≤ x≤a 0 ≤ x ≤ b2

M 1 = −Qx M 2 = − 12 Fx − Qa M 3 = − 12 Fx − Qa + F ( x − b2 ) M 4 = − 12 Fb + 12 Fb − Q ( a − x )

b 2

≤x≤b

0≤ x≤a

Applying Castigliano’s theorem, after setting Q=0, we have

δ h = EI1 ∫

1 2

δh =

Integrating,

Faxdx + EI1 ∫ ( − 12 Fax + 12 Fab)dx b2

Fab 2 8 EI

______________________________________________________________________________________ SOLUTION (10.10) a x

θ R

From symmetry,

0≤ x≤a 0 ≤ θ ≤ π2

M 1 = − Px M 2 = − Pa − PR sin θ

Hence, or

π 2

= EI1 ∫ M 1 ∂∂MP1 dx + EI1 ∫ a

π 2

0 2

δ = EI2 [ ∫ Px 2 dx + ∫

M 2 ∂∂MP2 Rdθ

P( a + R sin θ ) 2 Rdθ ]

= 6PEI ( 4a 3 + 6πRa + 24 R 2 a + 3πR 3 )

______________________________________________________________________________________ 263 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (10.11) P

A Pa/2

P/2

B a

M BD = P2 x

v D = EI1 ∫ M i ∂Pi dx ∂M

P/2

Pa/2

M AD = P2 ( x − a ),

v D = 2PEI [∫

2 a ( x − a )2 2

dx + ∫ x2 dx 2

Pa/2 Integrating, 3

v D = 4PaEI ↓

______________________________________________________________________________________ SOLUTION (10.12) Statics: FBC = 2 P3 , Thus

FCD = P3 ,

FAB = P3 ,

M AB = Px

BC AB U = ∫ 2FAE dx + ∫ M2 EIAB dx + ∫ 2FAE dx + ∫

1 = 2 AE ∫

P2 3

dx + 2 1RI ∫ P 2 x 2 dx + 0

D F2

C 2 AE 2L 1 4 2 AE 3 0

∫

1 P 2 dx + 2 AE ∫ P3 dx 2

We have

= ( 611AEL + 34EIL ) P 2

δ C = ∂∂UP = 3PLE ( 11A + 8 IL ) ↓ 2

______________________________________________________________________________________ SOLUTION (10.13) Referring to Fig. P10.13, we write

M = PR sin θ m = R sin θ

T = PR(1 − cos θ ) t = R(1 − cos θ )

Here t denotes torque caused by a unit load. Applying Eq. (10.13), deflection at the free end (perpendicular to the plane of the ring):

δ = EI1 ∫

π 2

( PR 2 sin 2 θ ) Rdθ + JG1 ∫

π 2

[ PR 2 (1 − cos θ ) 2 ]Rdθ π 2

sin 2θ sin 2θ θ θ = PR + PR 4 0 EI 2 − 4 0 JG θ − 2 sin θ + 2 + 3

= PR4 ( EIπ + 3πJG−8 ) 3

Letting J = 2 I = πr

δ =

PR 3 r4

2 , we have ( + ) 1 E

0.226 G

______________________________________________________________________________________ SOLUTION (10.14) Referring to Fig. P10.14:

M = − PR sin θ T = PR(1 − cos θ )

m = − R sin θ t = R(1 − cos θ )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.14 (CONT.) where t denotes torque caused by a unit load. We have and sin(2π − θ ) = − sin θ cos(2π − θ ) = cos θ Applying Eq. (10.13), deflection at the free end (perpendicular to the plane of the ring): 2π

2π

sin 2 θdθ + PR (1 − cos θ ) 2 dθ δ = PR EI ∫ JG ∫ 3

Integrating,

δ = PR 3π [ EI1 + JG3 ]

______________________________________________________________________________________ SOLUTION (10.15) Q

Segment AC

M 1 =−Qx

∂M 1 ∂Q =− x

Segment BC

∂M 2 p ( x − a)2 = −x 2 ∂Q a M ∂M 3 a M ∂M 1 1 2 2 Thus, = vA ∫ dx + ∫ dx 0 EI ∂Q a EI ∂Q M2 = −Qx −

Let Q=0. Hence

vA =

p 3a p 3a 3 2 x a xdx ( ) ( x − 2ax 2 + a 2 x)dx − = ∫ ∫ a a 2 EI 2 EI 3a

p x 4 2ax3 a 2 x 2 10 pa 4 = − + = ↓ 2 EI 4 3 2 a 3 EI

______________________________________________________________________________________ SOLUTION (10.16) B

L A

L= L= 2L AB CD L= L= L= L BC AC BD

The method of joints(or sections) is applied:

N AB =

Thus,

N BC = −W

N CD = − 2W N BD = W N AC = −W

∂N j 1 Nj Lj ∑ AE ∂W WL WL ↓ = [ 2( 2) 2 + 1 + 1 + 1 + 2(= 2) 2] 8.657 AE AE

δD =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.17) P x x’ A

Pa L

C L

Segment AB

a C M1 = −P x − x L L

Segment BC

M2 = − Px '− C

(a)

For this case C=0:

vC=

1 L Pax 1 a ax (− )(− )dx + − Px '(− x ')dx ' ∫ EI 0 L L EI ∫0 L

Pa 2 x 3 P x '3 Pa 2 = + = ( L + a) ↓ EIL2 3 0 EI 3 0 3EI (b)

∂M 1 ∂M 2 x = − , = −1. ∂C L ∂C

For C=0, we have :

θC=

x 1 L Pax 1 a (− )(− )dx + − Px '(− x ')dx ' ∫ EI 0 L L EI ∫0 L

Pa x3 P x '2 = + EIL2 3 0 EI 2 0 =

Pa (2 L + 3a ) 6 EI

______________________________________________________________________________________ SOLUTION (10.18) x” Q p A RA=Q/2 Segment AC

= M1 Let Q=0:

a x

C a x’

RB=2pa+Q/2

∂M 1 x 1 = Qx 2 ∂Q 2

∂M 1 1 a M1 dx = 0 ∫ 0 EI ∂Q

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.18 (CONT.) Segment CB

1 1 Q(a + x ') − Qx '− px '2 2 2 Q 1 = (a − x ') − px '2 2 2 ∂M 2 1 (a − x ') = ∂Q 2

M= 2

Let Q=0:

∂M 2 pa 4 1 1 a px '2 1 = − ' ( ) ( ') ' M dx = − a − x dx 2 48 EI 2 2 EI ∫ ∂Q EI ∫0

Segment BD

∂M 3 1 M3 = − px ''2 = 0 2 ∂Q

Therefore

= vC

∫ Mi

∂M i pa 4 = dx ↑ ∂Q 48 EI

______________________________________________________________________________________ SOLUTION (10.19) Q p L/2

pL 8

+ Q2

pL Q M1 = ( + )x 8 2

Segment CB

Let Q=0:

3 pL 8

+ Q2

∂M 1 x = ∂Q 2

3 pL Q px '2 + ) x '− 8 2 2

∂M 2 x ' = ∂Q 2

L 2 3 pLx ' pLx x px '2 x ' ( ) dx ' dx + − ∫0 8 2 ∫0 8 2 2 2 2 L 2 pLx L 2 3 pLx ' px '3 pL4 = p∫ dx + p ∫ ( − )dx ' = 0 0 16 16 4 96 4 pL = vC ↓ 96 EI

EIvC =

x’

Segment AC

M2 = (

L/2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.20) Consider RD as redundant. C 4

A 2.25 m

0: N ∑F = x

1.5 m

NAB

= AEδ D

∂N BD =1 ∂RD

∑F = 0: y

= N BC 1.25 P − 1.25 RD

= −0.75 P + 0.75 RD

∂N BC ∂RD = −1.25

where Thus,

NBD =RD

NBC

∂N AB ∂RD = 0.75

∂N j

0 N L ∑= ∂R j

= (1.25 P − 1.25 RD )(7.5)(−1.25) +(−0.75 P + 0.75 RD )(4.5)(0.75) + RD (3)1 =0 = −14.25 P + 17.25 RD = 0 or RD = 0.826 P Then N BD = 0.826 P N AB = −0.131P N BC = 0.22 P ______________________________________________________________________________________ SOLUTION (10.21) Let RA and M A be redundant. P

C L RA

1 lb

δP

1 lb ⋅ in.

m= −1(a + x)

M = − Px

vA =

m' = −1

Mm '

dx dx θ ∫ ∫= EI EI

v= AP

1 b Pb 2 a b ( Px )[ ( )] ( + ) a x dx − − + = EI ∫0 EI 2 3

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.21 (CONT.)

θ AP=

Pb 2 1 b Px dx ( )( 1) − − = EI ∫0 2 EI Let now RA and M A are applied. x

1 lb ⋅ in.

A RA

1 lb

m= x m' = 1

M= RA x − M A ( RA x − M A ) x 1 RA L3 M A L2 − = dx ( ) ∫0 EI EI 3 2 L ( R x − M )1 1 RA L2 A A θ AR ∫ = = − M A L) dx ( 0 EI EI 2 Since = v AP v= θ AP θ AR AR L

= v AR

Solving,

1 RA L3 M A L2 Pb 2 a b ( = ( ) + ) − 2 EI 2 3 EI 3 1 RA L2 Pb 2 ( = − M A L) 2 EI EI 3

= RA

Pb 2 (3a + b) ↑ L3

= MA

Pab 2 L2

______________________________________________________________________________________ SOLUTION (10.22) x p x A

R = kδ A

Refer to the above figures, we write:

1 M= Rx − px 2 2

m= −1x

Hence, deflection spring at A:

= δA

dx ∫= EI k

1 L 1 R ( Rx − px 2 )(− x= )dx ∫ 2 EI 0 k 3 pL 8 Solving R = 1 + (3EI kL3 ) =

Reactions at B may then be found from statics. ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.23) Consider RB as redundant. P

x A

Segment BC:

∂M 1 ∂RB = 0

M 1 = − Px

Segment AB:

M2 = − Px + RB ( x − a )

Thus,

∂M 2 ∂RB =− x a

EIvB =0 =∫ (− Px)(0)dx + ∫ [− Px + RB ( x − a )]( x − a )dx 3a

= ∫ [− Px 2 + RB x 2 − 2 RB ax + Pax + RB a 2 ]dx a

Px3 RB x3 Pax 2 = − + − RB ax 2 + + RB xa 2 3 3 2 a

from which

14 3 = − P + RB 3 8

= RB

7 P↑ 4

M = A

1 Pa 2

Statics:

= RA

3 P↓ 4

______________________________________________________________________________________ SOLUTION (10.24) Consider RA and M A as redundants. y P A a MA C L

b B

RA Segment AC:

M1 = RA x − M A Segment BC:

∂M 1 ∂M 1 = x = −1 ∂RA ∂M A

M 2 = RA x − M A − P ( x − a ) ∂M 2 ∂RA = x ∂M 2 ∂M A = −1

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.24 (CONT.) Thus,

EIv A == 0 ∫ ( RA x − M A ) xdx + ∫ [ RA x − M A − P( x − a )]xdx 0

=RA

Simplifying,

Similarly

a a L − a3 L2 − a 2 L3 − a 3 − M A + RA −MA −P 3 2 3 2 3 L2 − a 2 + Pa = 0 2

P 1 1 RA L3 − M A L2 − (3a + 2b)b 2 = 0 3 2 6 a

(1)

EIθ A = 0= ∫ ( RA x − M A )(−1)dx + ∫ [( RA x − M A − P( x − a)](−1)dx 0

= − RA

a L −a L2 − a 2 + M A a − RA + M A ( L − a) + P 2 2 2 − Pa ( L − a ) = 0

This reduces to

1 1 0 RA L2 − M A L − Pb 2 = 2 2

(2)

Solving Eqs.(1) and (2):

MA =

Pab 2 L2

R= B

Pa 2 (a + 3b) ↑ L3

Statics:

RA =

Pb 2 (3a + b) ↑ L3 M= B

Pa 2b L2

______________________________________________________________________________________ SOLUTION (10.25) Consider RB as redundant. MA

A RA

M0 L

B RB

M= − RB x + M 0

∂M ∂RB = −x

Therefore

L 1 1 2 0= EIvB = RB L3 − M 0 L2 ∫0 ( RB x − M 0 x)dx = 3 2 3 M0 Solving= RB ↓ 2 L

Statics:

3M RA = 0 ↑ 2 L

1 M A = M0 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.26) We introduce a couple moment C at point B. The expression for the moments are then,

0 ≤ x ≤ L2 L 2 ≤ x ≤ L

M 1 = − RD x M 2 = − RD x + P( x − L2 ) + C Applying Eq. (10.6), δ B = ∂U ∂P : L

δ B = EI1 {∫ [ − RD x + P( x − L2 ) + C ]( x − L2 )dx} L2

Setting C=0 and integrating,

δ B = − 48PLEI [ 1−( 325EI16kL ) − 2] 3

Similarly, applying Eq. (10.7), θ B = ∂U ∂C : L

θ B = EI1 {∫ [− RD x + P( x − L2 ) + C ]dx} L2

Setting C=0 and integrating,

15 PL 1 θ B = − 128 EI 1−( 3 EI kL ) 2

______________________________________________________________________________________ SOLUTION (10.27)

M = FR sin θ + PR(1 − cos θ ), Therefore

δv = 0 = EI1 ∫ M ∂∂MF dx

δ v = EI1 ∫ [ FR sin θ + PR(1 − cos θ )]( R sin θ ) Rdθ 0

πFR3

= 2 EI + 2 PR EI = 0,

F = − 4πP = 4πP ↑

______________________________________________________________________________________ SOLUTION (10.28) Apply Eq. (10.7) to write a

θ A = JG1 [ ∫ TA ∂∂TT dx + ∫ (TA − T ) ∂ (T∂T−T ) ] = 0 A

Integrating and simplifying:

TA a + (TA − T )b = 0

TA = Lb T

Condition of equilibrium gives

TB = T − TA = aL T

______________________________________________________________________________________ SOLUTION (10.29) We write

M 1 = Qx M 2 = Q ( a + R sin θ ) + PR(1 − cos θ ) Applying Eq. (10.6), with δ Q = 0 : ∂U ∂Q

0≤ x≤a 0 ≤θ ≤π

= 0 = EI1 ∫ Qx 2 dx + ∫ [Q ( a + R sin θ ) + PR(1 − cos θ )]( a + R sin θ ) Rdθ

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.29 (CONT.) Integrating, 3

QR 2 2 π PR 0 = QL 3 EI + EI (πa + 4 Ra + 2 R ) + EI (πa + 2 R ) 2

This may be written in the following form: 3

( πa + 2 R ) Q = 3 1 −πPR R R 2 π R 3

a [ + +4( ) + ( ) ] a 3 a 2 a

______________________________________________________________________________________ SOLUTION (10.30) po

RA MA

We have Then,

M = − RB x + pL0 x 2x 3x = − RB x + p06x

δ B = 0 = EI1 ∫ M ∂∂RM dx = EI1 ∫ ( RB x − p6 Lx ) xdx 0

This yields, after integrating,

RB = p100 L Then, from equations of statics: 2

R A = 410p0 L

M A = p150 L

______________________________________________________________________________________ SOLUTION (10.31) Moments are expressed by

0 ≤ x ≤ L2

M 1 = − Rx M 2 = − Rx + M 0

L 2

Applying Eq. (10.6), with δ R = 0 : ∂U ∂R

= 0 = EI1 ∫

≤x≤L

Rx 2 dx + EI1 ∫ ( − Rx + M 0 )( − x )dx L2

from which

R = 98ML0

Using Eq. (10.7), slope at C:

θ C = ∂∂MU = 0 + EI1 ∫ ( − Rx + M 0 )dx L2

5M 0L 64 EI

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.32) p E1I1

E2I2

x A

L1 VB

HC VC

0 ≤ x ≤ L2 0 ≤ x ≤ L1

M1 = H Ax M 2 = −V A x + H A L2 + 12 px 2

Applying Eq. (10.6), with δ Av = 0 and δ Ah = 0, respectively:

Letting

∂U ∂V A

= E11I1 ∫ ( −V A x + H A L2 + px2 )( − x )dx = 0

∂U ∂H A

= E21I 2 ∫ H A x ( x )dx + E11I1 ∫ ( −V A x + H A L2 + 12 px 2 ) L2 dx = 0

(a)

(b)

λ = EE II LL

2 2 1

1 1 2

Equations (a) and (b) become

8V A − 12 H A L2 = 3 pL12 3λV A L1 − 2(3λ + 1) H A L2 = λpL12

V A = 23((3λλ++14)) pL1 = R Av 2

1 H A = 4 ( 3λλ +4 ) pL L2 = R Ah

The remaining reactions may then be found by using the equations of equilibrium. ______________________________________________________________________________________ SOLUTION (10.33) MC

B x

θ R

P/2

a/2

Clearly, the problem is statically in determined. We have

0 ≤ x ≤ a2

M x = −M C M x = − M C + 12 PR (1 − cos θ )

Slope at C is zero: ∂U ∂M C

= EI1 ∫

π 2

0 ≤ θ ≤ π2

( − M C R + PR2 − PR2 cos θ )dθ + EI1 ∫ 2

a 2

− M C dx = 0

(CONT.) ______________________________________________________________________________________ 274 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 10.33 (CONT.) Integrating, 2

(π − 2 ) M C = PR 2 ( a +πR )

Hence, for 0 ≤ θ ≤ π2 : 2

(π − 2 ) PR PR M = − PR 2 ( a +πR ) + 2 − 2 cos θ

For 0 ≤ x ≤ a : 2

(π − 2 ) M = PR 2 ( a +πR )

Therefore, M is maximum along a, for θ = 0. ______________________________________________________________________________________ SOLUTION (10.34) ( a ) Introduce Q at point E, as shown. (disregard C/L’s). A

B L

P L

Q C L

Joint E

N EB N ED

C L

N ED = − 2 P (C )

E 1

N EB = 2 P + Q (T )

Similarly we obtain the remaining member forces. Joint B

Joint D

N BD = − P (C )

N DA = 2P 2

N BA = 2 P + Q (T )

N DC = −3P (C )

δE = ∑

N j L j ∂N j AE ∂Q

1 = AE [( 2 P )(5)(1) + ( 2 P )(5)(1) + 0 + 0 + 0 + 0] = 10AE2P

( b ) Introduce couple C/L applied perpendicular to line DE at points D and E as shown in the figure ( now disregard Q ). Joint E P E

N EB

N ED

N ED = −( P + C2L ) 2 (C )

N EB = P + 22CL (T )

1 C L

Similarly, Joint D

N DA = 2 2 P Thus

N DC = −3P − 22CL (C )

P P ( 22 + 22 + 2 + 62 ) = AE ( 102 + 2 ) θ DE = AE1 ∑ N j ∂C = AE ∂N j

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.35) The complementary energy for the ith member of length Li , from Eq. (2.39) is σi σ 3

U 0*i = ∫

dσ = 4σKi 3 = 4 K1 3 ( NAii ) 4

The complementary energy of the truss is thus 6

U * = ∑ 4 Kj 3j ( A jj ) 4 AL

j =1

Equation (10.16) is then 6

δ E = ∑ K ( A ) 3 ∂P Lj

∂N j

(P10.35)

j =1

Applying the method of joints, as needed, we obtain

N 2 = − 54P N3 = 0 N 5 = N 6 = − 34P

N 1 = 32P N 4 = 54P

We have L= 1

L= L= 3 m, L= 4 m, L= L= 5 m and Ai = A. 5 6 3 2 4

Equation (P10.35) becomes

δ E = ( AK1 ) 3{L1 N 13 ( ∂∂NP ) + L2 N 23 ( ∂∂NP ) + ⋅ ⋅ ⋅ ⋅ ⋅ + L6 N 63 ( ∂∂NP )} P 3 or = δ E ( AK ) {3( 32 )3 ( 32 ) + 5(− 54 )3 (− 54 ) + 0 1

+5( 54 )3 ( 54 ) + 3( 34 )3 ( 34 ) + 3(− 34 )3 (− 34 )} P 3 48×81+5×625×2 + 3×81×2 ) = ( AK 256 P 3 ) δ E = 41.5( AK

______________________________________________________________________________________ SOLUTION (10.36) C 41

A P Joint A

L AC = 6.403 m, LCB = 5 m L AB = 8 m N AC

B P/2

P/2

∑N = 0: ∑N = 0: y

N AC =

N AB = 83 P

P/2

N AB

41 8

N BC = − 85 P

Joint B

A horizontal unit load is applied a C:

n AB = 83

n BC = − 85

n AC =

41 8

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.36 (CONT.) Thus

δ C = AE1 ∑ n j N j L j P [( 841 )( 841 )(6.403) + (− 85 )(− 85 )(5) + ( 83 )( 83 )(8)] = AE P = 718 . AE →

______________________________________________________________________________________ SOLUTION (10.37) ( a ) Refer to Fig. (a) with no unit load:

M AB = 0 M BE = Rx M EC = Rx − P ( x − L) M CD = R( 2 L) − PL

C L

The vertical deflection at A is zero. Thus

E L

δ v = EI1 ∑ M i ∂∂MR dx = 0

Figure (a)

= ∫ 0dx + ∫ Rx ( x )dx + ∫ [ Rx − P ( x − L)] xdx + ∫ ( 2 LR − PL)( 2 L)dx = 0

Integrating, we have 29 R = 64 P

( b ) Introducing a horizontal unit load at A, Fig. (a), we write Hence

m AB = x

m BE = L

m EC = L

mCD = L − x

δ h = EI1 ∑ M i mi dx L

L 2L

= EI1 {∫ 0dx + ∫ Rx ( L)dx + ∫ [ Rx − P( x − L)]Ldx + ∫ ( 2 LR − PL)( L − x )dx} 0

Integrating, after substituting R = 29P 64 , results in PL δ h = 13 32 EI

______________________________________________________________________________________ SOLUTION (10.38) x

M0 B

x a

1N C R

Introduce a vertical upward unit force 1 N at point C. Then Segment CB:

M1 = Rx

Segment BA:

m1 = x

M 2 = Rb = M 0

m2 = b

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.38 (CONT.) Thus

0 b

EIδC = 0 = ∫ M1 m1 dx + ∫ M 2 m2 dx = ∫ Rx 2 dx + ∫ ( Rb − M 0 )( b)dx 0

0 2

= Rb + Rab − M 0 ba 1 3

3M a

R = b( 3a +0 b ) ↑ ______________________________________________________________________________________ SOLUTION (10.39) We have α = 6 5 (Table 5.2), dx = ds = Rdθ Let a downward unit load of 1 N in addition to load P is applied at the free end in Fig. P10.39. We write

N = P cos θ V = P sin θ M = PR(1 − cos θ )

Thus,

3π

n = cos θ v = sin θ m = R(1 − cos θ ) 3π

(a)

3π

1 δ v = EI1 ∫ MmRdθ + AG ∫ αVvdx + AE1 ∫ Nnds 2

Substitution of Eqs.(a) gives 3π

3π

PR δv = PR (1 − cos θ ) 2 dθ + 56AG EI ∫ ∫ sin θdθ + AEPR ∫ cos 2 θdθ 3

Integrating

PR PR + 34πAE ↓ δv = ( 94π + 2) PREI + 34.6AG

______________________________________________________________________________________ SOLUTION (10.40) P/2 1N A A A NA MA θ 1N m R C P/2 Actual loading

Dummy loading

Because of symmetry about the vertical axis, only one half of the circle need be analyzed. Referring to the figure above, we have the following moments: For 0 ≤ θ ≤ π2 :

M 1 = M A + N A R(1 − cos θ ) − 12 PR sin θ

m1 = R(1 − cos θ ),

m1' = 1

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.40 (CONT.) For π2 ≤ θ ≤ π :

M 2 = M A + N A R(1 − cos θ ) − 12 PR

m2 = R(1 − cos θ ),

(b)

m2' = 1

Horizontal deflection and slope are zero at A. Thus, from Eqs. (10.13) and (10.14): π

δ Ah = EI1 ∫ MR (1 − cos θ ) Rdθ = 0 0

∫ M (1 − cos θ )dθ = 0 π π θ = ∫ Mm dx = ∫ M (1)dx = 0

(c)

1 EI

∫ M = 0 , where M represents M and M .

(d)

Substituting Eq. (d) into Eq. (c), the latter reduces to π

∫ M cos θdθ = 0

(e)

Introducing Eqs. (a) and (b) into Eqs. (d) and (e), we have π

M A ∫ dθ + N A R ∫ (1 − cos θ )dθ − 12 PR ∫

π 2

sin θdθ − 12 PR ∫ dθ = 0 π 2

M A ∫ cos θdθ + N A R ∫ (1 − cos θ ) cos θdθ

and

− 12 PR ∫

π 2

Integrating and solving,

N A = 2Pπ

sin θ cos θdθ − 12 PR ∫ cos θdθ = 0 π 2

M A = PR 4

______________________________________________________________________________________ SOLUTION (10.41) From Eq. (P10.41), we have

U = GJ2 [( θaC ) 2 a + ( θbC ) 2 b]

Using Eq. (10.22): ∂U ∂θ C

Solving, Then and

= T;

GJθ C ( a1 + 1b ) = T

Tab θ C = GJL

TA = GJa θ C = Lb T TB = GJb θ C = aL T

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.42)

L= L= 4.24 m, L= 3m AD DC BD U = 12 ALEε 2 = 12 ALE (δ v cosLα ) 2

Vertical load of the joint, using Eq. (10.22), 3

= P = ∑ ELi Ai i δ v cos 2 α

∂U ∂δ v

i =1

= P EAδ v [ cosLAD45 + cosLDC45 = + L1BD ] 0.5689 AEδ v 2

Substituting the numerical values,

P = 355.56 kN

______________________________________________________________________________________ SOLUTION (10.43) p x L y The deflection curve may be expressed by Eq. (10.23) and the bending strain energy is given by Eq. (10.25). The strain energy of deformation of the foundation is (Chap. 9): ∞

U 2 = 12 k ∫ v 2 dx = 14 kL ∑ a n2 0

n =1

Work done by the uniform load:

∞

W = 12 ∫ pvdx = p( πL )∑ n1 a n cos nπLx 0 0

n =1

∞

W = 2πpL ∑ ann

n =1, 3,⋅⋅⋅

Then, principle of virtual work yields π 4 EI 2 L3

n 4 a n + kL2 a n = 2npL π 4

a n = nπ ( n 44π pL 4 EI + kL4 )

Substituting this into Eq. (10.23) we obtain the required deflection curve. ______________________________________________________________________________________ SOLUTION (10.44) P L x A y We obtain and

v ' ' = 3 aL13 ( L − x ) L

1 ( L − x ) 2 dx U = EI2 ∫ ( v ' ' ) 2 dx = 92EIa L6 ∫

3 EI 2 L3

2 1

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.44 (CONT.)

δW = P ⋅ δv A Thus, δW = δU gives, P ⋅ δa1 = 32EIL ( 2a1δa1 )

(b)

Also

v A = a1 = 3PLEI

______________________________________________________________________________________ SOLUTION (10.45) Strain energy, by Eq. (c) of Sec. 10.10:

U = π64EI a2 L3 4

Work done by the load P is L

W = ∫ pvdx = pa ∫ dx − pa ∫ cos 2πxL dx = pa ( L − 2πL )

Applying the Ritz method, we obtain dΠ da

Solving

= dad (U − W ) = π64EI − pL(1 − 2πL ) = 0 L3 4

a = 0.1194 pL EI

Substitution of this into Eq. (P10.45) and letting x=L result in the deflection v A at the free end of the beam. ______________________________________________________________________________________ SOLUTION (10.46)

W = P ⋅ v A = P ( a1 L2 + a 2 L3 ) = PL2 ( a1 + a 2 L) L

Thus,

U = EI2 ∫ ( v ' ' ) 2 dx = 2 EIL( a12 + 3a1a 2 L + 3a 23 L2 ) 0

∂Π ∂a1

= 0:

PL2 = 2 EIL( 2a1 + 3a 2 L)

∂Π ∂a2

= 0:

PL3 = 6 EIL2 ( a1 + 2a 2 L)

Solving, a1 = 2PL EI

a 2 = − 6PEI

Substituting back into Eq. (P10.46):

v = 6PxEI (3L − x ) 2

______________________________________________________________________________________ SOLUTION (10.47) c P B L We have

y L

Π = EI2 ∫ ( v ' ' ) 2 dx − P ⋅ v B

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 10.47 (CONT.) Here

v = ax ( L − x ) = axL − ax 2 v ' = aL − 2ax, v ' ' = −2a

Substituting these into Eq. (a), after integration, we obtain

Π = 2a 2 EIL − PacL + Pac 2

and

dΠ da

= 4aEIL − PcL + Pc 2 = 0

L−c ) a = Pc4(EIL

The deflection of the beam at point B is therefore 2

L−c ) v B = Pc 4(EIL

______________________________________________________________________________________ SOLUTION (10.48) The assumed deflection is of the general cubic form:

v = a1 x 3 + a 2 x 2 + a3 x + a 4

For the left hand portion of the beam: (at x = 0 ); v=0

(at x = L 2 );

v' = 0

(a)

v' = 0 v=∆

(at x = 0 )

(at x = L 2 );

(1, 2) (3, 4)

Here ∆ is, deflection at midspan, to be determined. Eqs. (1) and (2) give a 3 = a 4 = 0 Eq. (3) yields

a 2 = − 3a41L

Eq. (4) gives

a1 = − 16L3∆

Introducing these into Eq. (a):

v = 4 ∆L3x (3L − 4 x )

0 ≤ x ≤ L2

Strain energy is

U = 2( EI2 2 ) ∫

( v ' ' ) 2 dx + 2( EI2 ) ∫ ( v " ) 2 dx

(b)

(c)

∆ Inserting Eq. (b) into Eq. (c) and integrating, we obtain: U = 72 EI L3

We have

∆ Π = 72 EI − P∆ L3

The midspan deflection, from ∂Π ∂∆ = 0, is thus PL ∆ = 144 EI = v max 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (10.49) We obtain ∞

v ' ' = ∑ a n ( nLπ ) 2 cos nπLx n =1

EI 2

∞

∫ (v' ' ) dx = ∑ a ( π ) 2

EI 2

∞

n =1

W = P ( v ) x = L = P ∑ 2a n 2

2 n

n L

4 L 2

( n = 2, 4, 6, ⋅ ⋅ ⋅ )

From the minimizing condition, ∂Π ∂a n = 0, we obtain or

EIa n ( nLπ ) 4 L2 − 2 P = 0 ∞

a n = ∑ n44πPL4 EI Therefore,

( n = 2, 4, 6, ⋅ ⋅ ⋅ )

∞

1 v max = v ( L2 ) = π8 PL 4 EI ∑ n 4 3

( n = 2, 4, 6, ⋅ ⋅ ⋅ )

End of Chapter 10

CHAPTER 11 SOLUTION (11.1) Pcr 1

(c 4 − b 4 )=

Pcr

F 2 = Pcr 1

4 (254 − 20 ) 18.11×104 mm 4 =

Pcr π 2 EI π 2 (200 ×109 )(0.1811×10−6 ) = = = 143 kN n nL2 2.5(1) 2 = F 2.5(143) = 357.5 kN Justification of the formula used:

A = π (c 2 − b 2 ) = π (252 − 202 ) = 706.86 mm 2

and

181,100 = 5.06 mm 706.86

I = A

= r

L 1000 = = 197.6 O.K. r 5.06

______________________________________________________________________________________ SOLUTION (11.2) (a)

Same area:

π 4

(d o2 − di2 ) =bo2 − bi2

bi2 =bo2 − or

(d o2 − di2 ) =502 −

bi= 42.35 mm (b)

Circular bar

(d o4 − di4 ) =

= Pcr

π 2 EI

π 64

= L2e

π 4

(502 − 402 )

1 (bo − bi )= 3.83 mm 2

(504 − 404 ) =181×10−9 m 4

π 2 (72 ×109 )(181×10−9 )

= 14.29 kN

(3) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.2 (CONT.) Square bar

1 4 4 1 (bo − bi )= (504 − 42.354 )= 252.8 ×10−9 m 4 12 12 2 π EI π 2 (72 ×109 )(252.8 ×10−9 ) = = 19.96 kN Pcr = L2e (3) 2

______________________________________________________________________________________ SOLUTION (11.3)

LBC = 22 + 12 = 2.236 m

d 50 = = 12.5 mm 4 4

4 = d4 (40) = 125.7(10−9 ) m 4 , 64 64 Bar BC ( L r 2, = = 236 12.5) 179 . Thus

= I

( FBC )= all

Pcr π 2 (70)(125.7) = = 9.65 kN 1.8(2.236) 2 n

( FAB )= all

Pcr π 2 (70)(125.7) = = 24.1 kN 1.8(1.414) 2 n

Bar AB

Joint B F

FAB

LAB =

12 + 12 = 1.414 m

∑ F =0 : 2 F − 5 F =0, F =0.791F x

B 1

∑ F= 0 : 2 F + 5 F − F= 0, F= 1.061F y

FBC

Solving

The allowable value for F:

F = 1.061FAB

F = 1.341FBC

F < 1.341(9.65) = 12.94 kN F < 1.061(24.1) = 25.6 kN

Thus

Fall = 12.94 kN

______________________________________________________________________________________ SOLUTION (11.4)

Le 0.7 = L 0.7(0.9) = 0.63 m P= nP = 2.4(25) = 60 kN , = cr = I

= d4 A 64

Equation (11.7) gives

= d4

πd2

d = r 4 4

64 Pcr L2e 64(60 ×103 )(0.63) 2 , = = d 0.0223 m = 22.3 mm π 3E π 3 (200 ×109 )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.4 (CONT.) Hence

Le 0.63 = = 113 r 0.0223 4

Equation (a) of Sec.11.6:

Le E 200 ×109 = 88.86 < 113 (= )c π = π r σy 250 ×106 Euler formula is valid, Therefore

d = 22.3 mm

______________________________________________________________________________________ SOLUTION (11.5) Refer to Solution of Prob. 11.4. Now we= have Le

0.7(600) = 420 mm

and Pcr 2.4(130) = = 312 kN . Equation (11.7):

= d4 and

64 Pcr L2e 64(312 ×103 )(0.42) 2 , = = d 0.0275 m π 3E π 3 (200 ×109 )

Le 420 = = 61.1 < 88.86 r 27.5 4

Euler formula does not apply. Use Johnson formula, Eq. (11.11):

σ yp L2e 1 Pc 312 ×103 250 ×106 (0.42) 2 1 2 = + 2 ) 2 2[ d 2( = + ] πσ yp π E π × 250 ×106 π 2 (200 ×109 ) or d = 0.041 m = 41.0 mm

______________________________________________________________________________________ SOLUTION (11.6) 13 LBC = 0.52 m FAB = 125 P, FBC = 12 P

P B

FAB

12 13 5 F

Bar AB

( FAB= )cr Bar BC

= ( FBC )cr

π 2 EI

= L2

π π 2 (200×109 )[ (0.006)4 ] 4 (0.32)2

= 19.62 kN =

π π 2 (200×109 )[ (0.009)4 ]

= 37.62 = kN

4 (0.52)2

13 12

5 12

Pcr ,

= Pcr 47.1 kN

Pcr= , Pcr 34.73 kN

Choose the small value, Pcr = 34.73 with n = 2.3 . Thus

Pall=

Pcr n

= 15.1 kN

34.73 2.3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.7) Differential equation for a free-body of a segment of x is given by Eq.(11.1):

d 2v + Pv = 0 dx 2 d 2v v 0 p + p 2= = 2 dx

EI or

General solution is

P EI

= v A sin px + B cos px

(1)

Boundary conditions :

v(0) = 0 : 0 + 0 += B 0, = B 0 = v '( L) 0 : AP = cos pL 0,= AP 0 Since A ≠ 0 : π P P cos L = 0 or L = n( ) EI EI 2 2 π P For n 1,= = EI 4 L2 Therefore

Pcr =

π 2 EI 4 L2

______________________________________________________________________________________ SOLUTION (11.8)

1 3 (240 ×1203 − 190 × 70 ) 29.13 ×106 mm 4 = 12 A = 240 ×120 − 190 × 70 = 15.5 ×103 mm 2

I min =

I min= A 43.35 mm

= rmin

= Le 0.5 = L 4.5 m

= Le r 4500 = 43.35 103.8 Hence,

= σ cr

π 2E π 2 (200 ×109 ) = = 183.2 MPa ( Le r ) 2 (103.8) 2

______________________________________________________________________________________ SOLUTION (11.9)

= Le 0.7 = L 6.3 m . From solution of Prob.11.8: rmin = 43.35 mm . We now have

= Le r 6300 = 43.35 145.3

Hence,

= σ cr

π 2 E π 2 (200 ×109 ) = = 93.5 MPa ( Le r ) (145.3) 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.10) 30 kN B x 1.8 m 1.3 m A 1

FBD

0 : − 30(3.1) + 0, F (1.8) = ∑M = 2 A

FBD = 73.1 kN

4 4 = I b= 12 50= 12 520.8 ×103 mm 4 ,

A = 50 × 50 = 2.5 ×103 mm 2

I L 1800 = 14.4 mm = = 125 A r 14.4 π 2 E π 2 (210 ×109 ) So, = σ cr = = 132.6 MPa ( L r )2 (125) 2 (solution is valid) We have σ cr < σ pl

= r

( FBD )cr = 132.6 ×106 (2.5 ×10−3 ) = 331.5 kN ( FBD )cr 331.5 and = n = = 4.5 73.1 FBD

______________________________________________________________________________________ SOLUTION (11.11) A solid circular section of diameter d:

= A

π 4

= I

Using Eq. (11.13a), we have

π 64

= r

I d = A 4

4 Pcr 1 σ 4L = σ pl − ( pl e ) 2 2 πd E 2π d 2 2 Pcr σ pl Le 1 2 Q.E.D. Solving, ) = + d 2( πσ pl π 2 E A rectangular section of height h and width b:

= A bh = I By Eq. (11.13a), we obtain

1 3 h2 bh = r2 12 12

Pcr 12 Le 2 1 σ ) = σ pl − ( pl bh E 2π h

from which

Pcr 3L2σ hσ pl (1 − 2e pl2 ) π Eh It is assumed that h ≤ b . b=

Q.E.D.

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.12)

σ all=

P 90 ×103 = = 30 MPa A 3 ×10−3

(1)

2π 2 (200 ×109 ) = 125.7 250 ×106 Assuming L r ≥ Cc , use Eq. (11.8) and (1):

= Cc

π 2 (200 ×109 ) σ all = 2 = 30 ×106 , 1.92( L r )

L = 185 r

Our assumption was correct. Hence,

= L (185) = r (185)(25) = 4, 625 mm or L = 4.625 m ______________________________________________________________________________________ SOLUTION (11.13) P

2π 2 (210 ×109 ) = 121.7 280 ×106 Le 4.2 = = 42 r 0.1

Cc = 6m

Le=0.7L

Since L ry < Cc , use Eq. (11.9) and (11.8):

5 3 42 1 42 3 n= + ( )− ( ) = 1.79 3 8 121.7 8 121.7 1 − (42) 2 [2(121.7) 2 ] = (280 ×106 ) 147.1 MPa = σ all 1.79 So,

3 P= 147.1(27 ×10= ) 3972 kN all

______________________________________________________________________________________ SOLUTION (11.14) P

δα1

k 1

δα2

δα3

L L L

k 3

(CONT.) ______________________________________________________________________________________ 289 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 11.14 (CONT.) The moment expression about joints 1 and 2:

∑ M = 0; ∑ M = 0;

or, in general,

PLδα1 − k (δα1 − δα 2 ) = 0

PL(δα1 − δα 2 ) − k (δα 2 − δα 3 ) = 0

PL( ∑ δα i ) − kδα n = 0

(P11.14)

i =1

In matrix form, we have

 PL − k  PL   ⋅   ⋅  PL

PL − k ⋅ ⋅ PL

 δα1  0  δα  0   2    ⋅ ⋅ ⋅ ⋅ ⋅  ⋅  =  ⋅   ⋅ ⋅ ⋅ ⋅ ⋅  ⋅   ⋅      PL PL ⋅ ⋅ PL − k  δα n  0 0 k

0 0

⋅ ⋅ ⋅ ⋅

0 0

That is

[ A]{δα } = {0} Determinant A = 0, yields Pcr . In our case, n = 3.

______________________________________________________________________________________ SOLUTION (11.15) (a)

Pcr is proportional to I . We have 4

I s = π4r

15 πr 15 I h = π4 [ r 4 − ( 2r ) 4 ] = 16 4 = 16 I s Thus, reduction in Pcr is 6.25 % 4

( b ) Substituting the given data: 15π (0.015)4 64

= 37.276(10−9 ) m 4

= Ih Pcr =

π 2 EI h

= L2

π 2 (110×109 )(37.276×10−9 ) (1.5)2

= 17.99 kN

______________________________________________________________________________________ SOLUTION (11.16) ( a= ) Pcr

= I

2(100) = 200 kN Pcr L2

π 2E

200(103 )(2)2

= 7.369(106 ) mm 4

π 2 (11×109 )

6 a = = a 96.97 mm 7.369(10 ); 12 4

We have

σ=

P A

100(10)3

= 10.63 MPa <15 MPa

(0.09697)2

Thus, a cross section of 97 × 97 mm is acceptable. (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.16 (CONT.)

2(200) = 400 kN

( b= ) Pcr

400(103 )(2)2

4 = 14.738(106 ) mm= ; a 115.32 mm

= I

π 2 (11×109 )

We obtain

σ=

200(103 )

P A

= 15.04 MPa >15 MPa

(0.11532)2

Dimension is not acceptable. Therefore, 200(103 )

a 2= A=

15(106 )

;

a= 115.5 mm

Use a cross section of

116 ×116 mm

______________________________________________________________________________________ SOLUTION (11.17) We have 1(50)4 12

= 0.521×106 mm 4 .

= I Hence

Thus

= Pcr

π 2 EI

Pall=

Pcr n

Pall 2.5

π 2 (200×103 )(0.521)

= L2

(3)2

= 114.3 kN

= 52 kN

114.3 2.2

= 20.8 kN

52 2.5

______________________________________________________________________________________ SOLUTION (11.18) ( a ) Using Eq. (11.5),

= Pcr

π 2 (210×109 )(0.075×0.053 12)

127.5 kN = 0.49(3.6)2 (2)

127.5(103 )

= σ cr

= 34 MPa

3.79(10−3 )

( b ) We have

I min = 0.07512( 0.05) and Thus,

−4 r 2 = I min A = 2.08(10 )

Le r

Also,

( 3.6 ) = 0.7r L = 00.7.0144 = 175 1

C c = [ 2σπ ypE ] 2 = 121.673 2

As 121.673 < 175, use Eq. (11.13): π 2 (210×109 )

= σ all

= 35.25 MPa 1.92(175)2

Note that

σ=

P A

450(103 ) (0.05)(0.075)

= 120 MPa

and yielding does not occur. But member fails as a column, since Pcr < P.

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.19) y P P A C θ Me L Me D x L B P P θ Symmetrical buckling shown in the figure, creates relative bending moments M e which resist free rotation of the ends of the member AB and CD. Thus, for member AB: 2

EI ddx 2v = − Pv + M e

with the general solution

v = C1 cos λx + C 2 sin λx + MP2

where,

λ = 2

(a)

P EI

Boundary conditions are

v ( 0) = 0

v ' ( L2 ) = 0

v ' (0) = θ = M2 EIe L Introducing v from Eq. (a) into these: C1 + MPe = 0, − C1λ sin λ2L + C 2 λ cos λ2L = 0

C 2 λ = M2 EIe L

The foregoing lead to the following trancendental equation or

tan λ2L + 2PL λEI = 0 tan λ2L + λ2L = 0

from which λL 2 = 2.029. Thus, 2

EI Pcr = 16.47 = ( 0.π774EIL )2 L2

The effective length in the situation described is therefore equal to 0.774L. ______________________________________________________________________________________ SOLUTION (11.20) Let

λ1 = M λ2 = EI

P EI

The governing differential equation is

EIv ' ' = M − Pv

v ' '+ λ1v = λ22

y L

P=W

Solution is

v = A cos λ1 x + B sin λ2 x + ( λλ12 ) 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.20 (CONT.) Boundary conditions give:

v ' ( 0) = 0 = B v (0) = A + ( λλ12 ) 2 = 0;

A1 = −( λλ12 ) 2

v ' ( L) = ( λλ12 ) 2 sin λ1 L = 0

sin λ1 L = 0; Choose λ1 L = π . Then,

λ1 L = 0, π , 2π , ⋅ ⋅ ⋅

1 2

λ1 = πL = ( EIP ) ; Thus,

P = π LEI 2

Le = L

______________________________________________________________________________________ SOLUTION (11.21) For the vertical bar, from Eq. (11.5): 2

Pcr = π4 EI L

The midspan deflection of the beam is thus 4

P L 5 PL δ = 384 EI − 48 EI cr

2 5 PL 5 PL π L L δ = 384 EI − 192 = 192 ( 2 EI − π ) 4

______________________________________________________________________________________ SOLUTION (11.22) ( a ) We have α ( ∆T ) L = δ 2 or (b) where Thus,

∆T = 2δαL PL α ( ∆T ) L − AE = δ2 2

P = π4 LEI2 2

δ π I L 1 ∆T = 2δαL + π4 LEI2 AE αL = 2αL + 4 L2 Aα

______________________________________________________________________________________ SOLUTION (11.23) L/4 L/2 L/4 (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.23 (CONT.) For a fixed ended column Le may be determined as follows. Inflection points and midpoint divide the bar into 4 equal portions. Each portion is the same as the fundamental case (Fig. 11.2a). Thus Le = L 4 . For the dimension given, we obtain L = r

= 65.99

1.94 0.0294

Equation (11.6) gives then

σ cr = (4Lπ r E) = 0.0091E 2

Substituting E = 174(10 ),

σ cr 0.0091(175 ×= 109 ) 1592.5 MPa = Since, from Fig. P11.20, E is valid only up to 175 MPa, the 1592,5 MPa cannot be critical stress.

Similarly, for inelastic range, from Eq. (11.7):

σ cr = 0.0091Et1

= 0.0091(46.7 = ×109 ) 425 MPa Applying the same reasoning as before, this value is also not a critical stress. For E t 2 : 9 σ cr 0.0091(28 ×10 ) 254.8 MPa = =

We now observe from the sketch that 254.8 falls in the stress range for which E t 2 is valid. Thus, the buckling load:

254.8(0.0323) = Pt σ= = 823 kN cr A

______________________________________________________________________________________ SOLUTION (11.24) A P C

For bar AB: or So,

−5 1.41 = P 4.2(106 )(5.4 ×10 = ) 226.8 P = 160.4 N 160.4 = σ AB 5.4(10 = 2.97 MPa < 4.2 MPa −5 )

For bar BC: 2

−9

)(3.91×10 ) P = Pcr = π LEI = π (210×106.25 2 2

= 1297 N 1297 And = σ cr 5.4(10 = 24 MPa > 6.3 MPa −5 ) Conclusion:

Bar BC fails as a column, Pcr = 1297 N.

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.25) Referring to Fig. P11.25, we write

I y = 2 Arc2 + 2 Ad 2

= 2(1.719 ×10−3 )(0.0225) 2 + 2(1.719 ×10−3 )(0.0125 + 0.02325) 2 = 6.13(10−6 ) m 4 1

ry = ( 2 Ay ) 2 = ( rc2 + d 2 ) 2

and

= (0.02252 + 0.035752 ) 2 = 42.24 mm We see that ry for two channels is the same as for one channel. But rz=

rc= 0.0225 m. Thus, rz < ry . Columns tends to buckle with respect to z axis.

The slenderness ratios: Le rz

Le rz

.10 = 02.0225 ≈ 94,

From Sec. 11.7: 2

.20 = 04.0225 ≈ 187

( 210×10 ) = 20,420; C c2 = 2σπ ypE = 2π 203 (106 ) 2

( a ) In this case 0 < ( Le

C c = 143

r ) < C c . Then, substituting the given data,

the first of Eqs. (11.8) yields

σ all = 97.685 MPa

( b ) Now we have: C c ≤ ( Le

r ) ≤ 200 and the second of Eqs. (11.13) gives

σ all = 30.87 MPa

______________________________________________________________________________________ SOLUTION (11.26) From Fig. P11.25, we observe that

I y > Iz .

Therefore the column buckles with respect to the z axis. Since, L r

We obtain

.2 = 0.40225 = 186.7

π 2 (210×109 )

σ cr = = ( Lπ Er )

= 59.46 MPa (186.7)2

______________________________________________________________________________________ SOLUTION (11.27) From Eq. (11.6), 2

We have

(140×10 ) 2 ( Lr ) lim it = [ π 280 ] = 70.2 ×106

I = bh12 = Ar 2 3

I = ( 0.05)(120.025) = 1.25 × 10 −3 r 2 Solving, r = 7.216 mm Hence, ( Lr ) actual = 7.2161(.210− 3 ) = 166.3 or

(CONT.) ______________________________________________________________________________________ 295 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 11.27 (CONT.) Thus, elastic buckling occurs, since ( L r ) lim it < ( L r ) actual . We have ∆ = α (∆T ) L and ε = ∆L = α ( ∆T ) . The condition that

ε = α ( ∆T ) − ( Lπ r ) = 0 2

results in

∆T = α (πL r )2 = 10(10−6π)(166.3)2 = 35.7o C 2

______________________________________________________________________________________ SOLUTION (11.28)

σ all=

125(103 )

P A

= 40.85 MPa

3.06(10−3 )

Also 2

( 200×10 ) 2 ] = 125.7 C c = [ 2π 250 (106 )

Assuming ( L r ) ≥ C c : σ all = Solving, Lr = 158.6

π 2 ( 200×109 ) 1.92 ( L r ) 2

) = 1028( L.08r ()10 2

O.K.

Choosing the smallest radii of gyration: L ry

L = 0.0246 = 158.6

from which

L = 3.9 m

______________________________________________________________________________________ SOLUTION (11.29) P

= A π= (20) 2 1256.64 mm 2 I =

40 mm

vmax

e (a)

Pcr =

0.7 m B

(20) 4 125.66 ×103 mm 4 =

π 2 EI

L2 π 2 (200 ×109 )(125.66 ×10−9 ) = (0.7) 2 = 506.2 kN

Figure (a)

Using Eq. (11.18):

 π 60   o 1= = ×10−3 e sec   − 1 e sec(31 ) − 1 , 2 506.2    

e = 6 mm

(CONT.) ______________________________________________________________________________________ 296 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 11.29 (CONT.) (b)

Referring to Fig. (a):

M = P(vmax + e)= 80(1= + 6) 560 N ⋅ m

80 ×103 560(20)10−3 P Mc Hence, σ max = + = + 1256.64 ×10−6 125.66 ×10−9 A I = 63.66 + 89.13 = 152.8 MPa ______________________________________________________________________________________ SOLUTION (11.30)

2= L= L 1.4 m . Refer to solution of Prob. 11.29 e

π 2 EI

= Pcr (a)

π 2 (200 ×109 )(125.66 ×10−9 )

= L2e

(1.4) 2

= 126.6 kN

Equation (11.18):

 π 60   o 1= = ×10−3 e sec   − 1 e[sec(62 ) − 1] ,   2 126.6   (b)

e = 0.88 mm

M = P(vmax + e)= 80(1 + 0.88)= 150.4 N ⋅ m . Therefore,

σ max=

150.4(20 ×10−3 ) P Mc += 63.66 + = 63.66 + 23.9 = 87.6 MPa 125.66 ×10−9 A I

______________________________________________________________________________________ SOLUTION (11.31)

100 mmC 20 mm

200 mm D

B P

L= L 3.6 m 2= e A = 200 ×100 − 160 × 60 = 10.4 ×103 mm 2 1 (200 ×1003 − 160 × 603 ) 12 = 13.79 ×106 mm 4 = I

= r = I A 36.4 mm e= c= 50 mm Thus,

Le ec 50 × 50 = = 1.89 = 49.45 2 2 r (36.4) 2r

(CONT.) ______________________________________________________________________________________ 297 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 11.31 (CONT.) Use Eq.(11.19) with L = Le :

  250 ×103  250 ×103  + 1 1.89sec 49.45    = 98.7 MPa  10.4 ×10−3  70 ×109 (10.4 ×10−3 )     

= σ max

______________________________________________________________________________________ SOLUTION (11.32)

200 mm

A P

20 mm

100 mm B

L= 2= L 3.6 m e A = 200 × 100 − 160 × 60 = 10.4 ×103 mm 2 1 (100 × 2003 − 60 ×1603 ) 12 = 46.19 ×106 mm 4 = I

e= c= 100 mm

= r

= I A 66.6 mm

Therefore,

Le ec 100 × 100 = = 2.25= 27.03 2 2 r (66.6) 2r Apply Eq.(11.19) with L = Le :

= σ max

  250 ×103  250 ×103  + 1 2.25sec 27.03    = 85.7 MPa −3  9  × × 10.4 ×10−3  70 10 (10.4 10 )    

______________________________________________________________________________________ SOLUTION (11.33)

= I

1 12

π 2 EI

= Pcr e=

3 (50)(25) = 65.1×103 mm 4

= L2

h 2

π 2 (210)(65.1) (1.5)2

= 59.97 kN

= 12.5 mm,

25 2

P= 10 kN

Equation (11.18) gives 10 π = vmax 0.0125[sec( = 3.10 mm 2 59.97 ) − 1]

Thus

M max =P(e + vmax ) =10(12.5 + 3.10) =156 N ⋅ m

______________________________________________________________________________________ 298 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ SOLUTION (11.34) We have

= I

π 1 ( D 4 − d 4 ) and A = ( D 2 − d 2 ); r = I A = D 2 + d 2 4 4 64

= P n= (430) 1.4(430) = 602 kN ,

I =

(1902 − 1702 )= 5655 mm 2

(1904 − 1704= ) 22.97(106 ) mm 4

64 1 2 = r 1902 + 170= 63.7 mm 4 P 602 ×103 L 3.5 Hence, 106.5 MPa 55 = = = = −6 A 5655 × 10 r 63.7 × 10−3 ec 0.095e 23.41e = = 2 r (63.7 ×10−3 ) 2 Substitute these into Eq. (11.19b):

  106.5 ×106    = 210 106.5 1 + 23.41e sec  27.5  9   190 × 10    

from which

e = 0.033 m = 33 mm

______________________________________________________________________________________ SOLUTION (11.35) From solution of Prob. 11.34:

A = 5655 mm 2 I = 22.97(106 ) mm 4

Hence

vmax

Pcr =

π 2 EI

L2 π 2 (200 ×109 )(22.97 ×10−6 ) = (4.5) 2 = 2.239 MN

Figure (a)

( a ) Using Eq. (11.18):

 π = 1.2 e sec    2

50 ×103   =  − 1 e sec(13.45o ) − 1 , 2.239 ×106   

e = 42.55 mm

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.35 (CONT.) ( b ) Referring to Figure (a):

M = P(vmax + = e) 50(1.2 + 42.55) = 2188 N ⋅ m

Hence,

P A

Mc I

σ max== +

50 ×103 2188(0.095) + = 17.89 MPa −6 5655 ×10 22.97 ×10−6

______________________________________________________________________________________ SOLUTION (11.36)

= Le 2(5) = 10 m . Refer to solution of Prob. 11.35

π 2 EI

= Pcr

= L2e

π 2 (200 ×109 )(22.97 ×10−6 ) (10) 2

= 453.4 kN

( a ) Equation (11.18):

 π 1.2 e sec  =   2

50   o =  − 1 e[sec(29.89 ) − 1] , 453.4  

e = 7.82 mm

M = P(vmax + = e) 50(1.2 + 7.82) = 451 N ⋅ m. Therefore, P Mc σ max= + A I 50 ×103 451(0.095) = + = 8.842 + 1.865 = 10.71 MPa −6 5655 ×10 22.97 ×10−6 (b)

______________________________________________________________________________________ SOLUTION (11.37)

= p 77(103 )(0.05 ×= 0.05) 192.5 N m (0.05)4 12

= 5.2(10−7 ) m 4

= I (a)

σ max=

Mc I

192.5(9)2 0.025 8 5.2(10−7 )

= 93.705 MPa

5 pL v max = 384 EI

5(192.5)(9) = 0.1506 m = 150.6 mm

384(210×109 )(5.2×10−7 )

( b ) Taking

= a0

5 pL4 384 EI

= 150.6 mm

We obtain, using Eq. (11.15):

vmax =

0.1506 227.5 mm = 4500(9)2

1−

π 2 (210×109 )(5.2×10−7 )

Applying Eq. (11.16):

σ = max

4500 0.025

[1 + 0.2275 5.2(100.025 ] −7 ) 0.025

= 49.4 MPa ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.38) Given e=0.05 m, c=0.1016 m and

Iz rz2 == A

Then,

Pcr =

= 7.765(10−3 ) mm 2

45.66 5880

π 2 EI y

π 2 (210×103 )15.4

= L2

(4.5)2

= 1576 kN

π 2 (210×103 )45.66

= 4673 kN (4.5)2

Also, = Pcr

Thus, Eq. (11.19) becomes 6 = 210(10 )

P 5880(10−6 )

[1 + 0.05(0.1016) sec( π2 7.765(10−3 )

P 4673×103

)]

from which, by trial and error,

P ≈ 700 kN = Pmax

______________________________________________________________________________________ SOLUTION (11.39) Governing equation is

EIv1 ' ' = −( v0 + v1 )

where,

v0 = a1 sin πLx + 5a1 sin 2Lπx

This may be written We have

v1 ' '+ λ2 v1 = −λ2 a1 sin πLx − 5λ2 a1 sin 2Lπx

(a)

v1 = c1 cos λx + c2 sin λx + v p

(b)

Particular solution is

v p = A sin πLx + B cos πLx + D sin 2Lπx + E cos 2Lπx

(c)

Substituting Eq. (c) into Eq. (a): 2

A = πλ2 a1 (

B=0

− λ2 ) 2

D = 45πλ2 a1 (

Thus,

E =0

− λ2 )

v p = (π λL )a21−λ2 sin πLx + 4 (π5λL )a21−λ2 sin 2Lπx

Boundary conditions

v1 (0) = 0 yield c1 = c2 = 0.

v1 ( L) = 0

General solution is

v = v1 + v p = v0 + v p

Letting

b = λπ2 L2 = πPL2 EI 2

we obtain

v = 1a−1b sin πLx + 204−ab1 sin 2Lπx

Solution of the b is found from

v ( 34L ) = 0 = 1a−1b sin

π(

3L ) 4 L

+ 20 1a−1b sin

3L ) 4 L

2π (

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.39 (CONT.)

Thus,

b = 0.89 2 b = πPL2 EI ,

and

P = 0.89 π LEI 2

P = bπL2EI 2

______________________________________________________________________________________ SOLUTION (11.40) x ds dv dx v p(x)

P M

V+dV P M+dM dv dx

+ ⋅⋅⋅

An element isolated from the beam is shown in a deformed state in the figure above. Assume sin θ ≈ θ , cos θ ≈ 1, and ds ≈ dx. Then

∑F = 0:

− V + pbx + (V + dV ) = 0

p = − dV dx

(a)

∑M = 0:

M − Pdv − Vdx + pdx dx2 − ( M + dM ) = 0

Neglecting terms of second order, this becomes dv V = − dM dx − p dx

(b)

If the shear and axial deformations are neglected, the moment at any point is 2

M = EI ddx 2v

(c)

Substituting of Eqs. (c) and (b) into Eq. (a) gives

( EI ddx 2v ) + P ddx 2v = p For EI = constant, 2 d 2v + EIP ddx 2v = EIp dx 2 d2 dx 2

(d) (e)

Homogeneous solution of this equation is

v = c1 sin

P EI

x + c2 cos

x + c3 + c 4

P EI

where, c1 through c4 will require for evaluation, four boundary conditions. ______________________________________________________________________________________ SOLUTION (11.41) Given

v = a0 [1 − ( 4Lx2 )] 2

8 xa

v ' ' = − 8La20

and hence, v ' = − L2 0

Potential energy function is

Π = 2∫

1 2

EI ( v ' ' ) 2 dx − 2 ∫

64 EIa0 64 Pa0 12 L L3

Thus,

δΠ δa 0

Pcr = 12LEI 2

1 2

32 Pa0 0 − 12 P( v ' ) 2 dx = 32 EIa L L3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.42) Assume v = v0 sin( πLx ). Then,

U = 2∫

EI 1 (1 + Lx2 )( v ' ' ) 2 dx

= 2 EI 1v0 ' ' πL4 [ ∫ 4

sin 2 πLx dx + L2 ∫

x sin 2 πLx dx ]

= 2 EI 1v02 πL4 [ L4 + π2 L2 ( π16 + 14 )] 4

We also have L

(a)

2 2

vπ 2 2 2 ∫ (v' ) dx = v0 πL2 ∫ cos πLx dx = 20 L 2

Thus,

∫ EI (v' ' ) dx = P ∫ (v' ) dx 2

yields

Pcr = π LEI2 1 ( 23 + π22 ) = 1.7 π LEI 2 2

______________________________________________________________________________________ SOLUTION (11.43)

v ' ' = 2 Lv12

We have v ' = 2v1 Lx2

Potential energy function is L

Π = EI21 ∫ (1 − 2xL ) 4Lv41 dx − P2 ∫ 4Lv41 x 2 dx = 23 EIL13v1 − 23 v1LP Hence, gives

δΠ δv1

0 3 EI1v1

−

4 v1P 3L

Pcr = 94EIL21

______________________________________________________________________________________ SOLUTION (11.44) Given deflection curve equation satisfied the boundary condition of the column at x=L. The total potential energy, by Eq.(g) of Example 11.5, becomes:

= Π

EI 2

∫ ( υ ) dx − ∫ 0

d2 2 dx 2

P 0 2

( ddxυ ) 2 dx

= 6a 2 L3 ( EI − 52 PL2 ) + abL2 (6 EI − 52 PL2 ) + 2b 2 L( EI − 13 PL2 )

Equation (10.34) is therefore ∂Π ∂a

= 12 L3 ( EI − 52 PL2 )a + L2 (6 EI − 52 PL2 )b = 0

∂Π ∂b

= L2 (6 EI − 52 PL2 )a + 4 L( EI − 13 PL2 )b = 0

A nontrivial solution of this column exits when determinant of the coefficients is zero: 2

Solving,

P + 240 ( EIL4) = 3P 2 − 104 EI 0 L2

EI = P1 2.486 = Pcr L2

P2 = 32.181 EI L2 This critical load of the column is solely 0.76% larger than the exact value of π EI 4 L . ______________________________________________________________________________________ 2

______________________________________________________________________________________ SOLUTION (11.45) ∞

∑ a sin( π ). Then, U = π ∑n a

Assume v =

n x L

4 EI 4 L3

and

2 n

W = 12 ∫ P( v ' ) 2 dx + ∫ pvdx 0

= 4 L ∑ n a + π ∑ a n n1 Hence, from δU = δW , by letting 2 b = PPcr = πPL2 EI 2

π 2P

we obtain

2 PL

∞

1 a n = π4 PL 5 EI ∑ n 3 ( n 2 −b ) 4

1, 3, ⋅ ⋅ ⋅

Thus,

∞

1 sin nπLx v = π4 PL 5 EI ∑ n 3 ( n 2 −b ) 4

1, 3, ⋅ ⋅ ⋅

______________________________________________________________________________________ SOLUTION (11.46) 2F

L/2

L/4

Let c1 = L4 and c2 = L2 . Deflection curve is expressed by ∞

v = ∑ a n sin nπLx

(a)

U = EI2 ∫ ( v ' ' ) 2 dx

and

= F ∑ a n sin nπLc1 + 2 F ∑ a n sin nπLc2 + P2 ∫ ( v ' ) 2 dx

Therefore,

∂ (U −W ) ∂an

gives or

π 4 EI 2 L3

n 4 a nδa n = Fδa n sin nπLc1 + 2 Fδa n sin nπLc2 + P2πL a n n 2 2

a n = F sin(πn2πEIn4 )2+2 F sin(P nπ 2 ) 2 L3

( n2 −

Pcr

)

Solution is found by substituting this into Eq. (a). ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (11.47) P x L

Boundary conditions are: v (0) = v ( L) = 0 We have Thus,

Hence,

v = aLx 2 + bLx 3 − ax 3 − bx 4

v ' = 2aLx + 3bLx 2 − 3ax 2 − 4bx 3 v ' ' = 2aL + 6 BLx − 6ax − 12bx 2 L

Π = EI2 ∫ ( v ' ' ) 2 dx − P2 ∫ ( v ' ) 2 dx 0

EI 2

[4a L + 8abL + 245 b 2 L5 ] − P2 [ 152 a 2 L5 + 15 abL6 + 353 b 2 L7 ] 2 3

It is required that ∂Π ∂a

= EI2 (8aL3 + 8abL4 ) − P2 ( 154 aL5 + bL5 ) = 0

∂Π ∂b

= EI2 (8aL4 + 485 bL5 ) − P2 ( 15 aL6 + bL35 ) = 0

Letting λ = PL EI , these become 2

( 4 − 235λ )a + ( 4 − 10λ )bL = 0 ( 4 − 10λ )a + ( 245 − 335λ )bL = 0 Since a ≠ 0 and b ≠ 0 : − ( 4 − 10λ ) 2 + ( 4 − 215λ )( 245 − 335λ ) = 0 or

Solving, Hence,

λ2 − 128λ + 2240 = 0

λ1 = 20.9

λ2 = 107.1

Pcr = 20.L9 EI = 2.12 π LEI 2 2

______________________________________________________________________________________ SOLUTION (11.48)

L/2 EI1

L/2

0 y

L/3

EI2 L/3

L/3

From Eq. (11.29):

vm +1 + ( λ1h 2 − 2)vm + vm −1 = 0

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.48 (CONT.) Here

λ12 = EIP

λ22 = EIP

Applying Eq. (a) at 1 and 2:

Thus, or

λ1h 2 − 2 1   v1  0    =   2 h λ 1 − 2 2   v2  0

( λ1h 2 − 2)( λ2 h 2 − 2) − 1 = 0

λ1λ2 h 4 − 2λ1h 2 − 2λ2 h 2 + 3 = 0

This is written as

P 2 [ 81EL2 I I ] − P[ 92EIL1 + 92EIL 2 ] + 3 = 0 4

1 2

Solution is

( P1, 2 ) cr = 9LE2 ( I 1 + I 2 ) ± 9LE2 [ I 12 − I 1 I 2 + I 22 ] 2

When I 1 = I 2 :

Pcr = 9LEI2

______________________________________________________________________________________ SOLUTION (11.49) L/2 0

L/2 1

y Boundary conditions yield

v ' (0) = 0;

v0 = 0

v (0) = 0; v ' ' ( L) = 0;

v1 = v −1 v3 − 2v 2 + 2v1 = 0

(1)

v ' ' ' ( L) = 0;

v 4 − 2v3 + 2v1 = 0

(2)

Then, applying Eq. (l) of Example 11.8 at points 1, 2, 0, respectively:

(7 − 2λ2 h 2 )v1 + ( λ2 h 2 − 4)v2 + v3 = 0

(3)

( λ h − 4)v1 + (6 − 2λ h )v 2 + ( λ h − 4)v3 + v 4 = 0

(4)

( λ2 h 2 − 4)v1 + v 2 = 0

(5)

From Eq. (5):

v 2 = −v1 ( λ2 h 2 − 4)

Equation (1) becomes

v3 = 2v1 − v1 = −2v1 ( λ2 h 2 − 4)v1 − 2v1 ( λ2 h 2 − 4) − v1

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.49 (CONT.) Substituting this into Eq. (3),

(7 − 2λ2 h 2 )v1 − ( λ2 h 2 − 4)( λ2 h 2 − 4)v1 − 2v1 ( λ2 h 2 − 4) − v1 = 0

from which

λ4 h 4 − 4λ2 h 2 + 2 = 0 λ2 h 2 = 0.59 = PhEI 2

or Thus,

Pcr = 2.36L2EI

______________________________________________________________________________________ SOLUTION (11.50) L 1 2 3 P P x 4

y From symmetry: v1 = v3 .

We have

I ( x ) = (1 + 2Lx ) EI 1 I ( x ) = (3 − 2Lx ) EI 1

0 ≤ x ≤ L2 L 2 ≤ x ≤ L

Equation (11.29), gives at 0, 1, 2, 3:

v1 + v −1 = 0; v1 = −v −1 2 v2 + [ ( 3 Ph 2 ) EI1 − 2]v1 = 0 v1 + [ Ph EI1 − 2]v 2 + v1 = 0 2

[ ( 3 Ph 2 ) EI1 − 2]v1 + v 2 = 0 2

The foregoing equations lead to

 23 λ2 h 2 − 2   v1  0 1    =   1 2 2 v 2 0 2 λ h − 2  2   from which, since v1 ≠ 0 and v 2 ≠ 0 : ( λ2 h 2 − 6)( λ2 h 2 − 1) = 0 Thus,

Pcr = 16 EIL21

______________________________________________________________________________________ SOLUTION (11.51) L/4 P

3L/4

EI1 0

EI2 1

y We have

α1 = LL 44 = 1

α 2 = LL 24 = 2.

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 11.51 (CONT.) Applying Eq. (11.30) at 1 and 2:

[ 16PLEI1 − 2]v1 + v 2 = 0

(a)

v1 [ 16PLEI1 − 1]v 2 = 0

(b)

2 3

Let

k1 = 16PLEI1 − 1 2

k 2 = 16PLEI 2 2

Equations (a) and (b) yield then

k1 − 2

1 =0 k2 − 1

2 3

from which or

( k1 − 2)( k 2 − 1) − 23 = 0 L P 2 [ 256 EL2 I I ] − P[ 162EI + 16LEI1 ] + 43 = 0 2 2

1 2

Solution of this quadratic equation gives the critical load as follows: 1

Pcr = 8 E ( 2LI21 + I 2 ) − 8LE2 [4 I 12 + I 22 − 1.24 I 1 I 2 ] 2 In a special case, for I 1 = I 2 = I , the preceding reduces to

Pcr = 24LEI − 15.L32EI = 8.7 EI 2 L2

End of Chapter 11

CHAPTER 12 SOLUTION (12.1) Components of stress are

σ x = PA + MrI

3375(0.05) ×10 =π90(0.05) =45.84 MPa 2 + π (0.05)4 4 3

τ xy= and

Tr J

4500(0.05)

= 22.92 MPa

π (0.05)4 2

σ y = σ z = τ xz = τ yz = 0

Thus, from Sec. 2.15 (in MPa):

 2σ3 x  τ xy 0 

τ xy

− σ3x 0

0  30.56 22.92 0    0  = 22.92 − 15.28 0    − σ3x   0 0 − 15.28

______________________________________________________________________________________ SOLUTION (12.2) A

a B

α D P At instability, member AD (or DC) and BD become in length:

L' AD = L' DC = LAD + 1−n1n1 LAD = 1L−ADn1

L' BD = LBD + 1−nn2 2 LBD = 1L−BDn2 Therefore, Initially:

2 a 2 + ( 1L−BDn2 ) 2 ( 1L−ADn1 )=

(a)

L2AD= a 2 + L2BD

(b)

Eliminating a from Eqs. (a) and (b), we obtain

= ( 11−−nn12 ) cos α = LLBD AD For We have

n1 = 0.2

n1 ( 2 − n1 ) n2 ( 2 − n2 )

n2 = 0.3

and 1

cos α = 00..78 [ 00..23((11..78)) ] 2 = 0.7351 or

α = 42.68o

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.3) p PL We write x 2 1 1 2

PL2

M ( x ) = 2 p( L − x )

Equation (12.7) is then 1

v ' ' = ( KIMn ) n = ( 2 KIP n ) n ( L − x ) n

v ' = − λ2( L− x2)

2 +1 n

+ c1

( +1)( + 2 ) n n 2 +2

λ ( L− x ) n

v= 2

+ c1 x + c2

2 ( +1)( + 2 ) n n

(a)

Boundary conditions yield 2 +1 n

v ' (0) = 0;

v (0) = 0;

c1 = λ2L n

2 +2 n

c2 = − 2 λL 2

Substituting these and Eq. (g) of Sec. 12.6 into Eq. (a), we obtain 2 +2 n

v = ( 2 Kp−n ) n [ 2( L− x )2

( +1)( + 2 ) n n

2 +2

− 2 Ln 2

2 +1

( +1)( + 2 ) n n

For n = 1, K = E , and x = L :

( +1)( + 2 ) n n

+ L2n x ] n

v = 6pLEI − 24pLEI = 8pLEI ______________________________________________________________________________________ SOLUTION (12.4) P We write x y

Hence,

M = P( L − x )

v ' ' = ( KIPn ) n ( L − x ) n = λ ( L − x ) n

v ' = − λ ( L1− x ) n

1 +2

λ ( L− x ) n

v= 1

1 +1 n

1 ( +1)( + 2 ) n n

+ c1

+ c1 x + c2

(a)

Boundary conditions yield

v ' (0) = 0;

1 +1 n

v (0) = 0;

c1 = λ1L n

1 +2 n

c2 = − 1 λL 1

Substituting these onto Eq. (a), we find an expression for the deflection.

( +1)( + 2 ) n n

For a special case of n = 1 and K = E , the deflection at free end:

v ( L) = 2PLEI − 6PLEI = 3PLEI 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.5) 1

We have σ = Kε 4 and I = 23 bh . Here 3

a = ε max h

ε = hy ε max = ay ,

Moment is thus,

M = ∫ σydA = ∫ yK ( ay ) 4 bdy = 89 Ka 4 −h

1 4

But

σ max = Kε max = Ka h

Hence,

M = ( 89 )bh 2σ max

1 4

σ max = 89bhM = 34MhI 2

______________________________________________________________________________________ SOLUTION (12.6)

0≤ x≤a

M = − Px Hence,

1 n

v1 ' ' = ( − KIPxn ) = −λx , 1 +2 n

v1 = − 1 λx 1

( +1)( + 2 ) n n

v1 ' = − λ1x n

1 +1 n

+ c1

+ c1 x + c2

(a)

Similarly, M = − Pa , at a ≤ x ≤ ( L − a ) : 1

v 2 ' = −λa n x + c3

v2 ' ' = ( − KIPan ) n = −λa n , 1

v 2 = −λa n x2 + c3 x + c4 2

Boundary conditions yield,

(b)

v1 (0) = 0;

c2 = 0

v 2 ( L2 ) = 0;

c3 = λa n L2

v1 ' ( a ) = v2 ' ( a );

c1 = λa n L2 − λa1

1 +1 n

n ( +1) n

1 +2 n

v1 ( a ) = v2 ( a );

1 +2

c 4 = λa 2 − 1 λa n 1

( +1)( + 2 ) n n

1 +2

− λa1n

n ( +1) n

Substitution of these of these constants into Eqs. (a) and (b) gives the required solution. For a special case of n = 1 and K = E , the midspan deflection is Pa v ( L2 ) = v max = PaL 8 EI − 6 EI 2

v max = 24PaEI (3L2 − 4a 2 )

We compute

y = 23.19 mm (measured from top surface) and I = 4.4(10−7 ) m 4 .

Therefore

) [3(1.2) 2 − 4(0.45) 2 ] v max = 24 ( 4.04.×4510(−8000 7 ) 200×109

= 0.00598 = m 6 mm ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.7)

AAB =

= ABC

π 4

2 (0.06) 2827.44 ×10−6 m 2 =

2 (0.05) = 1963.5 ×10−6 m 2

(a) Since ABC < AAB :

Pmax = σ yp ABC = 240 ×106 (1963.5 ×10−6 ) = 471.2 kN (b) Loading Segment AB deforms elastically:

δ= AB

PL 471.2 ×103 (1.5) = = 1.19 mm AE 2827.44 ×10−6 (210 ×109 )

Segment BC deforms plastically:

(δ BC ) max =δ AC − δ AB =10 − 1.19 =8.81 mm

After unloading Segment BC remains elastic. Thus

(δ AB ) p = 0

Segment BC remains plastic. We have:

= δ yp ε= yp L

σ yp

240 ×106 (1) 1.14 mm = = L 210 ×109 E

P Pmax

δyp

δBC

(δBC)p (δBC)max Referring to the above figure:

(δ BC ) p (δ BC ) max − δ yp = 8.81 − 1.14 = 7.67 mm =

______________________________________________________________________________________ SOLUTION (12.8)

= AAB

= ABC

π 4

2 (0.03) = 706.86 ×10−6 m 2

2 (0.02) = 314.16 ×10−6 m 2

(CONT.) ______________________________________________________________________________________ 312 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 12.8 (CONT.) (a) Since ABC < AAB :

280 ×106 (314.16 ×10−6 ) Pmax = σ yp ABC = = 87.96 kN (b) Loading Segment AB deforms elastically:

87.96 ×103 (1.5) PL = = 0.89 mm δ= AB AE 706.86 ×10−6 (210 ×109 ) Segment BC deforms plastically:

(δ BC ) max =δ AC − δ AB =10 − 0.89 =9.11 mm

After loading Segment AB remains elastic. Thus

(δ AB ) p = 0

Segment BC remains plastic. We have:

= δ yp

σ yp

280 ×106 = = (1) 1.33 mm L 210 ×109 E

Referring to the solution of Prob. 12.7:

(δ BC ) p (δ BC ) max − δ yp =

= 9.11 − 1.33 = 7.78 mm ______________________________________________________________________________________ SOLUTION (12.9) B

A L

40 40 o

Using Eqs. (P12.9):

C o

N= N= AD CD

= 0.309 Pu

P cos 2 40o 1 + 2 cos3 40o

= N BD

P = 0.527 P 1 + 2 cos3 40o

Since N BD > N AD , we have

0.527 Pu = σ yp A = 250 ×106 (400 ×10−6 ) or

Pu = 189.8 kN

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.10) Introducing the data into Eq. (g) of Example 2.5, we find that the axial forces in segments AC and CB of the bar are

= RA

P 1+[(0.6×400) (0.9×800)]

= RB

3P 4

P 4

The normal stresses are therefore

− 4(8003×P10 ) = −937.5 P σa = −6

σb =

= 625 P

P 4(400×10−6 )

where the minus sign indicates compression. Inasmuch as σ a = σ b , the load Pyp at the onset of yielding is obtained from

σ yp = 410 MPa . At this load, AC just yields, and the strain has the value of ε yp σ yp= E 2158 µ . So, = P= yp

σ yp

= 437.3 kN

937.5

δ= δ= ε yp= a 1.3 mm b a

(a)

As the road P further increases, segment AC becomes plastic, and the stress is uniformly equal to σ yp , thus resisting a compressive force of 328 kN . On the RA σ= = yp Aa other hand at the start of yielding, segment CB carries a tensile load of

= RB σ= 164 kN and the strain just reaches ε yp = σ yp E . Thus, yp Ab Pu = RA + RB = 492 kN

δ C = ε yp b = 1.94 mm

(b)

The volume of load is the ultimate load of the stepped bar. ______________________________________________________________________________________ SOLUTION (12.11) We have

ε= yp (a)

σ yp

= E

= 1000 µ

210 210(103 )

Loading. When σ is reached, the additional strain ∆ε of the bar equals

∆ε t=

= 417 µ

260 − 210 120(103 )

The total strain in the bar is then

ε= ε yp + ∆ε= 1000 + 417 = 1417 µ t t

(a)

The elongation of the bar ∆L :

∆Lt= L0ε t= 150(0.001417)= 0.21 mm

(b)

Unloading. Referring to Fig. P12.11,

ε= ε= εe p t

where= εe

(b)

= 1238 µ

260 210(103 )

(c)

Inserting Eqs.(a) and (c) into (b), we find

ε p =1417 − 1238 =179 µ

Hence, residual elongation on unloading equals

∆L= L0ε = 150(0.000179)= 0.03 mm e p

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.12) See Eqs.(a) & (b) of Example 12.4:

= Ps' 4.32 Pa' = 2 Pa' + Ps' 400

(1, 2)

Solving,

= Pa' 63.3 = kN Ps' 189.9 kN Then

= σs

189.9 ×103 63.3 ×103 = = 253 MPa σ = 126.6 MPa a 750(10−6 ) 500(10−6 )

Since 253 MPa > 240 MPa and 126.6 MPa < 320 MPa, steel bar yields while aluminum bars remain elastic. Thus

(σ a ) res = 0

(σ s ) res = 253 − 240 = 13 MPa

______________________________________________________________________________________ SOLUTION (12.13) Rod begins to yield at:

( Pr ) yp (σ (250)(45) = = = 11.25 kN r ) yp Ar

(δ r )(= ε r ) yp L

(σ r ) yp 250 ×106 = = L (1.2) 1.5 mm Er 200 ×109

The result is shown in Fig. (a). Here Yr corresponds to the onset

of yield in the rod.

Tube begins to yield at:

( Pt ) yp (σ (310)(60) = = = 18.6 kN t ) yp At

(ε t ) yp 310 ×106 = = L (δ t ) yp = (1.2) 3.72 mm Et 100 ×109 The result is shown in Fig. (b), where Yt represents the onset of yield

in the tube. Total P-δ of the rod-tube combination:

P= Pr + Pt

δ= δr = δt

The result is in Fig. (c).

P (kN)

Pt (kN) Pr (kN) 11.25

29.85

18.6

3.72

δt (mm) 0

Yt Yr

Yt 1.5

(a)

δr (mm) 0

1.5

(b)

3.72 δ (mm)

1.5

(c)

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.14) We have e = 6 − 2 = 4 mm, I = (1 12)(12)

= 1.728 ×103 mm 4 and c = 6 mm.

( a ) Applying Eq. (12.9):

M yp = or

σ yp I

= 350×61.728

= M yp 100.8 N ⋅ m

( b ) Equation (12.10) results in

M u = aσ yp ( a 2 − e3 ) = (0.012)(350)(12 2 − 43 ) 2

= 582.4 N ⋅ m ______________________________________________________________________________________ SOLUTION (12.15)

σ yp

σ yp c

h-c

Referring to this figure,

(σ yp ) ca2 = ( h − c )aσ yp

or Hence,

c = 2h3 M = 12 2 3ha σ yp 23 23h + h3 aσ yp h6 11 = ( 54 )ah 2σ yp

______________________________________________________________________________________ SOLUTION (12.16) p

L/2

C L

L/2

pL2 2

Deflection at C (Case 6 of Table D.4), in elastic range: 4

v max = 384pLEI Start of yielding. From Case 4 of Table D.5:

p = p yp = Thus, M L

12 M yp L2

( 2 bh 2σ

3) L2

σ L2

v max = 32ypEI = 32 E ( 2ypbh3 3) = 32ypEh

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.17) Equation (12.9): (a)

(b)

2 M yp = bh 2σ yp 3 2 × 0.06(0.04) 2 6 (240 ×10 ) 15.36 kN ⋅ m = = 3 Equation (12.10b):

3 1 20 2 (15.36 ×103 )[1 − = ( ) ] 21.12 kN ⋅ m 2 3 40

______________________________________________________________________________________ SOLUTION (12.18) (a)

Equations (12.9) and (12.10b):

3 1 e M = 1.3 = [1 − ( ) 2 ] 2 3 h M yp or

1 e2 0.867 = 1 − ( 2 ), 3 h

e= 0.632h

(b) The residual stress pattern will be as in Fig. 12.16c. ______________________________________________________________________________________ SOLUTION (12.19)

A = 2(10)(40) + (10)(30) = 1100 mm 2 Neutral axis divides section into two equal areas:

Solving

A (30)(10) + 2(10)(h1 ) = = 550 mm 2 2

= h1 12.5 = mm h2 27.5 mm Therefore,

Z= where

= y1

A( y1 + y2 ) 2

y 10

h1 z

h2 10

∑ A y= 1 [2(h )(10)( h ) + 10(30)(h − 5)=] 6.93 mm 2 ∑A A 2 i

= y2

h 1 = [2(10)(h2 )( 2 )] 13.75 mm A2 2

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.19 (CONT.) So

1100 (6.93 + 13.75) = 11,374 mm3 2 Mp = σ yp Z = (260 ×106 )(11,374 ×10−9 )

= Z

= 2.96 kN ⋅ m ______________________________________________________________________________________ SOLUTION (12.20) y

Thus,

f =

π c3

= S 2 4 2 π c 4c Z 2= Ay 2 = 2 3π 3 4c = 3

4c y= 3π

C z

= A

π c2

Z 16 = ≈ 1.7 S 3π

______________________________________________________________________________________ SOLUTION (12.21) y A1 h

h/2

b/2

7bh 2 bh h bh h − ]= 32 2 4 2(4) 8 I 2 1 b h S = [bh3 − ( )3 ] = h 2 h 12 2 2

= Z 2 A1 y1 − 2= A2 y2 2[

bh 2 1 15 (1 − ) = bh 2 6 16 96

Thus,

= f

Z 7bh 2 96 = = 1.4 S 32 15bh 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.22)

1 (70)(120)3 12 = 10 ×106 mm 4 I M yp = σ yp c 10 ×10−6 = (250 ×106 ) 0.06 = 41.7 kN ⋅ m

120 mm

z 70 mm

= M p fM = yp

3 (41.7) = 62.6 kN ⋅ m 2

Elastic rebound stress

Mc 62.6 ×103 (0.06) = I 10 ×10−6 = 376 MPa

σ '= max

The results are sketched (in MPa) below.

-250

116

376 Mu

Mu =0 -250

250

60 mm 250 Loading

-376

-116

Unloading

Residual stresses

______________________________________________________________________________________ SOLUTION (12.23) Initial Yielding:

σ yp = πNr + 4πMr 1 2

where

(a)

1 3

N yp = πr 2σ yp ,

M yp = σ ypπr 3 4

We express Eq. (a) in the from N1 N yp

+ MMyp1 = 1

(1)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.23 (CONT.) Fully Plastic Deformation: A2 A1

dy y

From a Mathematics Handbook table: 1

A2 = π2r − [e( r 2 − e 2 ) 2 + r 2 sin −1 ( er )] 2

A1 = e( r 2 − e 2 ) 2 + r 2 sin −1 ( er ) Thus,

N 2 = 2σ yp [e( r 2 − e 2 ) 2 + r 2 sin −1 ( er )]

(2)

We also write 1

dA = 2( r 2 − y 2 ) 2 dy r

Q = ∫ 2 y ( r 2 − y 2 ) 2 dy = 23 ( r 2 − e 2 ) 2 e

Here Q is the first moment of the area A2 . Hence, 3

M 2 = 2Qσ yp = 43 ( r 2 − e 2 ) 2 σ yp

(3)

M u = 43 r σ yp = 316π M yp 3

Solving Eq.(3), 2

e = [ r 2 − ( 43 σMyp2 ) 3 ] 2 Substituting this into Eq. (2): 2

N 2 = 2σ yp [ r 2 − ( 43 σMyp2 ) 3 ] 2 ( 43 σMyp2 ) 3 + 2r 2σ yp sin −1 [1 − ( 43r 2 σMyp2 ) 3 ] 2 The foregoing results in πN 2 2 N yp

= ( 316π ) 3 ( MMyp2 ) 3 [1 − ( 316π ) 3 ( MMyp2 ) 3 ] 2 + sin −1 [1 − ( 316π ) 3 ( MMyp2 ) 3 ] 2

(4)

The governing equations for yielding to impend and for fully plastic deformation are given by Eqs. (1) and (4). A sketch of these, interaction curves, are shown below. N N yp

1.0

0.8 0.6 0.4 0.2 0

r M

Initial yielding

Fully plastic N M yp N yp N

= 201

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

B M M yp

Figure P12.23 ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.24) Let P=N. Then referring to Fig. P12.14, we have Thus,

M = Nd = N (0.05 + 0.05 + 0.025) = 0.125N N M = 1 0.125

Also,

M yp N yp

and

(πr 3 4 )σ yp

= 0.025 4

πr 3σ yp

M M yp = 20( N N yp )

Now referring to Fig. P12.14, we find that B (1.68, 0.084) . Therefore,

N 2 = 0.084 N yp = 0.084(πr 3σ yp )

= 0.084π (0.025) 2 ( 280 × 10 6 ) = 46.18 kN ______________________________________________________________________________________ SOLUTION (12.25) The plastic hinges for at 1 and 2, Fig. (a). x L− x

θ Pu 1

2 x

θ Figure (a)

L-x

Apply the principle of virtual work:

Pu ( xδθ ) = M u [2(δθ + Lxδθ − x )]

from which

Pu = x ( 2LL− x ) M u

The condition dPu

dx = 0 gives x = L 2 . Minimum magnitude of the ultimate load is thus

Pu = 8 ML u

______________________________________________________________________________________ SOLUTION (12.26) b

σ yp

h1 e b1 2

h1 ≤ e ≤ h σ yp

0 ≤ e ≤ h1

b1 2

Figure (b)

Figure (a)

Figure(c)

Initial Yielding:

σ yp = NA + MI c = 2 ( bhN−b h ) + ( 2 3)(Mbh h−b h ) 1

1 1

1 3

3 1 1

(a)

(CONT.) ______________________________________________________________________________________ 321 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 12.26 (CONT.) Here,

N yp = 2(bh − b1h1 )σ yp

M yp = 32h (bh 3 − b1h13 )σ yp Equation (a) may now be written N1 N yp

+ MMyp1 = 1

(1)

Fully Plastic Deformation: For 0 ≤ e ≤ h1 (Fig. b):

N 2 = 2 A1σ yp = 2(b − b1 )eσ yp

from which

N2 2 ( b −b1 )σ yp

Figure (d)

We have (Fig. d):

A2 = b( h − h1 ) + ( h1 − e)(b − b1 ) = bh − eb + eb1 − b1h1 h − h1

b ( h − h1 )( +1) 2 2 2 2 Ay 2 y=∑ = = 1 bh −b1h1 −e b+ e b1

Thus,

bh − eb + eb1 −b1h1

M 2 = 2σ yp A2 y = (bh 2 − b1h12 − e 2 b 2 + e 2 b1 )σ yp

(b)

M 2 = M u = (bh − b h )σ yp . 2

e = 0:

For

2 1 1

Substituting the given data, the preceding expressions become

N yp = 1.68h 2σ yp ,

M yp = 0.922h 3σ yp

M u = 1.113h 3σ yp and

Mu M yp

= 1.21,

M yp = 0.55hN yp

Hence, Eq. (b) leads to M2 1.21M yp

= 1 − 3.116 ( NNyp2 ) 2

(2a)

This is valid for

0 ≤ 0 ≤ h1

0 ≤ NNyp2 ≤ 1.67

For h1 ≤ e ≤ h2 (Figs. c and e):

N 2 = 2[0.7h + 0.2h + ( e − 0.7)2h ]σ yp = ( −2.52h 2 + 4eh )σ yp

e = 4 hNσ2yp + 0.63h

M 2 = 2h( h − e)2( e + h 2−e )σ yp

= 2h σ yp − 2he σ yp 3

Figure (e)

Hence, M2

2 h 3σ yp

N2 = 1 − he 2 = 1 − 16 hN4σ2 2 − 0.h315 − 0.397 2 σ

1 M2 2.17 M yp

= 0.603 − 0.176(

N2 2 N yp

) − 0.529( NNyp2 )

(2b)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.26 (CONT.) which is valid for

h1 ≤ e ≤ h

1.67 ≤ NNyp2 ≤ 1

Equations (1) and (2) are the governing expressions of the plastic bending. A sketch of these, interaction curves, are shown in the figure given below. N N yp

1.0

0.8 0.6

Fully plastic

Initial yielding

0.4 0.2 0

0.2 0.4 0.6 0.8 1.0 1.2

M M yp

Figure P12.26 ______________________________________________________________________________________ SOLUTION (12.27) P 1

v 3

θ1

L/3

Lθ1 = Lθ 2 and δv = Lδθ1 = 23 Lδθ 2 from which 2δθ 1 = δθ 2 .

We have v =

4 3

2 3

θ2

2L/3

4 3

Applying the principle of virtual work:

Puδv = M uδθ 1 + M u (δθ 1 + δθ 2 ) + M uδθ 2 4 3 Pu Lδθ 1 = 6 Mδθ 1

Pu = 92MLu

Solving,

______________________________________________________________________________________ SOLUTION (12.28) x p p px (a) L L/2 L/2 e 1

3 θ3

θ1 2 θ2 1

Figure P12.28

(b)

v 4

θ4

(c)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.28 (CONT.) We have two different modes to be checked for collapse loading. Mode A, Fig. P12.28b:

θ1 = θ ( Le −e ) ;

θ1 + θ 2 = Lθe

Applying the principle of virtual work: e

δ ∫ (θ1 x ) Lx pdx + δ ∫ [θ1e − θ 2 ( x − e)] Lx pdx = M uδθ1 + M uδθ 2 + M uδθ 2 or

0 e

δ ∫ θ1 x Lx pdx + δ ∫ [θ1e − θ 2 ( x − e)] Lx pdx = M uδθ 2 ( Le + 1)

(a)

Integrating the left hand side of this equation, carrying out the algebra and simplifying:

δθ 2 p[ L −6 e ] = M uδθ 2 ( Le + 1) 2

Solving, p = e ( L −ue )

(b)

Then, dp de = 0 gives,

0 = − e26(ML−ue ) + e 6( LM−ue ) ;

e = L2

Introducing this value of e into Eq. (b), the collapse load is

Pu = 24LM2 u Mode B, Fig. P12.28c: From symmetry θ 3 = θ 4 = θ . Principle of virtual work gives: L

δ ∫ [θ L2 − θ ( x − L2 )] pdx = 3M uδθ L2

δ ∫ [θL − θx ] pdx = 3M uδθ L2

Integrating, 2

δθ L8p = 3M uδθ or

Pu = 24LM u

Note that the collapse load is the same for modes A and B. ______________________________________________________________________________________ SOLUTION (12.29) Assume that plastic hinges force at 1 and 2, as shown in Fig. (a). On the average, plastic limit load a bθ

pu 1

θ a

aθ

Figure (a)

Note that pu goes through a virtual displacement of aδθ 2 . Thus, applying the principle of virtual work:

( pu L) 12 aδθ = M uδθ + M u (1 + ab )δθ

(CONT.) ______________________________________________________________________________________ 324 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 12.29 (CONT.) Solving, 2 L−a pu = 2 ML u ( La ) −a 2

The unknown distance a is determined from dpu

da = 0. In so doing and

simplifying the result, we obtain

a = (2 − 2 ) L and b = L − a = ( 2 − 1) L. ______________________________________________________________________________________ SOLUTION (12.30) σ yp y x dy y h h x 3

σ yp

b ( a ) We have M yp = σ yp I

From geometry, x b2

Then,

= ( h h2 )2− y ;

I = 2∫

x = b ( h2−h2 y )

( 2 x )dy ( y 2 ) = bh48

Hence, total yielding moment

M u = 2 bh48 h1 σ yp = bh24 σ yp 2

Also,

M u = 12 (area of rhombus) × (distance between centroids) (σ yp ) = bh4 ( h3 )σ yp = bh12 σ yp 2

Thus,

Mu M yp

=2 y

(b)

r b 4

I = π ( b 4−a ) σ

M yp = π4 (b 4 − a 4 ) byp = 4πb (b 4 − a 4 )σ yp Also, referring to the figure: π

M x = 2 ∫ ∫ r 2 sin θdrdθ = 43 (b 3 − a 3 )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.30 (CONT.) Hence,

y c = MAx = 43 (π 2b)(−ba2 −a 2 ) = 34π bb2 −−aa 2 3

We therefore have, Mu M yp

(π 2 )( b 2 − a 2 )σ

= (π 4 b )( b4 −a 4 )σyp c = 163πb bb4 −−aa 4 yp

______________________________________________________________________________________ SOLUTION (12.31) Applying the principle of virtual work:

L/2 L

δ ∫ (θ x) pdx = 4M u (δθ )

Integrating,

MB or

pL2 4

(δθ ) = 4 M u (δθ ) 2

M u = pL 16

______________________________________________________________________________________ SOLUTION (12.32) (a)

Applying principal virtual work in each case.

W P

(b)

a θ

θ 2θ

P θ

2θ

θ θ

Pa (δθ ) = 4 M u (δθ )

Wa (δθ ) = 4 M u (δθ )

6M u Pa (δθ ) + Wa (δθ ) =

or Pa M u = 4

or Wa M u = 4

or 6 Pa M u + Wa M u =

Results are plotted below. Wa/Mu 6 4

b c

2 0

d 4

6 Pa/Mu

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.33) (a)

Refer to Example 12.11 Fig.12.21b:

Fig.12.21c:

or PL M u = 8

or WL M u = 8.8

W= ( L 2)(δθ ) (1.2 M u )(δθ ) + 2 M u (δθ )

P( L 2)(δθ ) = 4 M u (δθ )

Fig.12.21d:

P( L 2)(δθ ) + W ( L 2)(δθ ) = 2(1.2 M u )(δθ ) + 4 M u (δθ )

or 12.8 PL M u + WL M u = (b)

Results are plotted below. WL/Mu

12.8 8.8 a

b c

d PL/Mu 8 12.8 0 4 ______________________________________________________________________________________ SOLUTION (12.34)

γ max L

6000 ×10−6 (0.5) = 0.10 rad φ = (a)= c 0.03 τ yp 180 ×106 = = 2600 µ < 6000 µ γ= y 70 ×109 G and shaft is yielded.

From similar triangles:

6000 µ

ρ0

2600

2600 µ

60 mm

2ρ0

(b)

30 6000

ρ0 = 13 mm

Elastic core. Use Eq.(b) of Sec. 12.9:

T1=

πρ03 2

τ yp =

(0.013)3 (180 ×106 )= 621 N ⋅ m

(CONT.) ______________________________________________________________________________________ 327 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 12.34 (CONT.) Outer part. Use Eq.(c) of Sec. 12.9:

2π 3 2π (c − ρ03 )τ yp = (0.033 − 0.0133 )(180 ×106 ) 3 3 = 9.35 kN ⋅ m T2 =

Thus,

T = T1 + T2 = 9.97 kN ⋅ m

______________________________________________________________________________________ SOLUTION (12.35)

140 MPa (Table D.1) τ yp =

G= 26 ×106 GPa, (a)

For partially plastic shaft, using Eq.(12.19):

(

ρ0 3 c

) =− 4

3T 6T 4 =− Typ π c3τ yp

Substituting the given values

ρ0

6(4.5 ×103 ) )3 = 4− 0.0711 = 0.025 π (0.025)3 (140 ×106 ) ρ0 = 10.4 mm

(

(b)

= γy

= φ

ρ0φ

= L

τ yp G

= φ

τ yp L G ρ0

140 ×106 (1.2) = 0.6213 rad = 35.6o 9 26 ×10 (0.0104)

______________________________________________________________________________________ SOLUTION (12.36) 80 mm

a=2 m

50 mm

b=1.5 m

τ yp

240 ×106 = 3000 µ 80 ×109 cφ 0.04(0.25) (γ max ) AC = = = 5000 µ a 2 0.025(0.25) (γ max= )CB = 4167 µ 1.5

γ= yp

= G

Both segments are yielded and partially plastic. (CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ γmax 12.36 (CONT.) Segment AC

ρ0 =

cγ y

γ max

γy

ρo 0.04(3000 ×10−6 ) = 24 mm −6 5000 ×10 3 πc π (0.04)3 Typ= τ yp= (240 ×106= ) 24.127 kN ⋅ m 2 2

Use Eq. (12.19):

TAC = Segment BC

= ρ0 Typ=

π c3 2

4 1 24 (24,127)[1 − ( )3 ] = 30.43 kN ⋅ m 3 4 40

cγ yp 0.025(3000 ×10−6 ) = = 18 mm 4167 ×10−6 γ max

τ yp=

π (0.025)3 2

Use Eq. (12.19):

(240 ×106= ) 5.89 kN ⋅ m

4 1 18 TBC = (5.89)[1 − ( )3 ] = 7.12 kN ⋅ m 3 4 25

Total applied torque is therefore

T = TAC + TBC = 37.55 kN ⋅ m

______________________________________________________________________________________ SOLUTION (12.37) For r=0, Eq. (f) of Sec. 12.11 leads to c1 = 0. Then, in plastic zone 2 2

σ r = σ yp − ρω3 r ,

σ θ = σ yp

If the plastic zone extends to radius c: 2 2

σ c = σ yp − ρω3 c

which may be found directly from Eq. (12.30) by setting a=0. The outer elastic zone is represented by an annular disk yielding at the inner radius c, wherein radial stress is σ c . We follow a procedure similar to that described in Sec. 12.11 for an annular disk. Boundary conditions:

(σ r ) r =c = σ c ,

( u ) r =0 = 0

are substituted into Eqs. (8.37) to obtain c1 and c2 . We then determine the stresses in the elastic region as follows: σ yp

σ r = 24 N [3(1 + ν ) − (1 + 3ν ) rcb ](1 − br ) 2

σ yp

2 2

σ θ = 24 N [ bc (1 + br )(1 + 3ν ) + 3ν (3 + ν ) − 3(1 + 3ν ) br ] 2

where

c b ) −1] N 2 = 8+(1+3ν )[( 24

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.38) h a b The sand volume is

V = 12 (b − a )ah + 2( 13 a 2 h2 )

Slope

τ yp = (ah2 )

The ultimate torque is thus 2

Tu = ( 3b−6a ) a τ yp The yield torque given in Table 6.2, by letting τ = τ yp . ______________________________________________________________________________________ SOLUTION (12.39)

a 3

a 3 3

h 2a

3 ) = 33 ha 2

Volume = 13 h ( 12 ⋅ 2a ⋅ a

Slope = a h3 3 = a3h3 = τ yp ,

h = a 3 3 τ yp

Tu = 2V = 23 a 3τ yp

(a)

T ( b ) From Table 6.2: τ A = 20 aT3 = 20 8a3 1

Thus,

T yp =

8a3 20

τ yp

( c ) Referring to the preceding results in items (a) and (b): Tu T yp

= 53

______________________________________________________________________________________ SOLUTION (12.40) Equilibrium condition, from Eq. (8.2): d dr

(tσ r ) − t (σ θ r−σ r ) = 0

Profile is, using Eq. (8.36) with s = 1, rt = at a . For full plasticity

σ θ − σ r = σ yp Thus,

tσ r = ∫ atra

σ yp r

σ at

dr = − ypr a + c1

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.40 (CONT.) Since, (σ r ) r =b = 0;

c1 =

σ yp ata b

Then, noting that σ r = − pi at t = t a : pi

σ yp

= a ( 1r − 1b )

pi = a ( brb− r ) σ yp

______________________________________________________________________________________ SOLUTION (12.41) We have α = 1 2 and substituting the given data into Eq. (12.42):

t0 =

0.606(14×106 )(0.5)

0.606 pr

0 = (2 K 3 )( n 3 )n

(2×900×106 1.73)(0.2 1.73)0.2

= 0.0063 = m 6.3 mm ______________________________________________________________________________________ SOLUTION (12.42) In this case, we have σ z > σ θ .

The total force is where

P = 2πrtσ 1 σ1 = σ z σ 2 = σθ

(a)

The values of r and t are given by Eqs. (g) and (f) of Sec. 12.12. Substituting these into Eq. (a):

P = 2πr0 e −ε1σ 1

At instant of stability,

dP = ( ∂∂σP1 )dσ 1 + ( ∂∂εP1 )dε 1 = 0 dσ 1 dε 1

= σ1

Equations (12.28) has the form

σ 1 = f (α )ε 1n

from which stability condition is

ε1 = n

Then, Solving,

n = ( σK1 ) n (α 2 − α + 1)( 2−2α )

σ 1 = K ( 2n ) n ( α −1α +1 ) 2

( 1− n ) 2

( 2−1α ) n

(b)

Since ε 1 + ε 2 + ε 3 = 0, the maximum principal strain is

ε 1 = ln LL = ln tt + ln rr 0

Minimum strain is

ε 3 = ln tt ;

t = t0 ln −1 ε 3

Thus,

σ 1 = 2πPrt = 2πr ln εP t ln ε o

−1

2 0

−1

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 12.42 (CONT.) Substituting Eqs. (12.38b) and (12.38c) into this equation:

σ1 = t0 =

2πr0t0 ln −1 [( 2πr0σ 1 ln −1 [(

σ1 1 2 )

(α 2 −α +1)

σ1 1 2

(α 2 −α +1)

)

where, σ 1 is given by Eq. (b).

1− n α ( −1)] 2 2

Results of the preceding expressions are simplified by setting α = 1 2 . ______________________________________________________________________________________ SOLUTION (12.43) Since the mean radius does not change, we have dε 0 = 0. We also take σ 3 = σ r = 0. Material is incompressible dε L = dε 3 , or ε L = −ε 3 . The Levy-Mises equation is thus dε L σ L −σ θ

Hence, And

= dσεθL

σ L = 2σ θ

σ e = 23 σ L ε e = ( 23 )dε L = −( 23 )dε 3

Radius r0 = constant. Therefore,

P = 2πr0 ⋅ tσ L

At instant of instability, dP = 0 : dσ L

σL

Hence,

= − dtt = −dε 3 = dε L

ε e = 23 n = 23 ε L σ e = K ( 23 ) n = 23 σ L ,

t = t0 e − n

At instant of instability, the load is then given by Eq. (P12.43). ______________________________________________________________________________________ SOLUTION (12.44) ( a ) Use Eq. (8.10), with τ max = σ yp σ yp 2n

= bp2 i−ba 2 =

pi a 1−( ) 2 b

2n :

;

Substituting the given data:

50 =

60 50 2 ) b

1−(

5 6

= 1 − ( 50b ) 2

Solving, b = 122.5 mm ( b ) Apply Eq. (12.47) with k = 1 and n = 3 : σ

pu = nyp ln( ab );

50 − 60 = 250 2.5 ln( b )

− ( 150 )

= e 250 = 0.549 Solving, b = 91.11 mm or

50 b

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.45) Apply Eq. (12.61a): with c = 1.4a and k = 23 . We have pi = σ r at r = a. Thus 2

pi = 23 ( 260)[ln( 1.a4 a ) − ( 2 a 2) (−2(a1).24 a ) ] = 300.2[ −0.336 − 0.255] = −177.4 MPa ______________________________________________________________________________________ SOLUTION (12.46) ( a ) Using Eq. (8.11):

= p yp

= 21.39 MPa

420 36 − 25 3 2(36)

We have k = 1, and thus, σ

6 pu = nyp ln ab = 420 3 ln 5 = 25.53 MPa

(b)

( nyp ) 2 = σ θ2 − σ θ σ r + σ r2 or

+ 25 2 36+ 25 (140) 2 = p 2yp [( 36 36− 25 ) + 36− 25 + 1]

Solving, p yp = 22.92 MPa Now k = 2

3 , and hence, = pu 2= (25.53) 29.48 MPa 3

______________________________________________________________________________________ SOLUTION (12.47) Using Eqs. (12.57) and (12.58) with k = 1 and r = 0.25 m, 0.3 = pu 400 = ln( 0.2 ) 162.2 MPa 0.3 σr = −400 ln( 0.25 ) = −72.93 MPa 0.3 σ θ =400[1 − ln 0.25 ] =327.1 MPa and σ z = 12 (σ r + σ θ ) = 127.1 MPa

Unloading from pu . At r = 0.25 m, Eqs. (8.12), (8.13), and (8.20): 2

0.2 (162.2) 0.3 σr = −57.09 MPa (1 − 0.25 )= 0.3 − 0.2 2

σ = θ

0.2 (162.2) 0.32 − 0.22

0.3 (1 + 0.25 316.6 MPa = 2 )

0.2 = σ z 162.2 = 129.8 MPa 0.32 − 0.22 2

Residual stresses at r=0.25 m:

(σ θ ) res. = 327.1 − 316.6 = 10.5 MPa

(σ r ) res. =−72.93 − (−57.09) =−15.84 MPa (σ z ) res. = 127.1 − 129.8 = −2.7 MPa

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (12.48) We have k = 1. Refer to Eq. (12.57). Inner cylinder, at r = b :

− pb = − pu + (σ yp ) i ln ab Outer cylinder, at r = c : 0 = − pb + (σ yp ) o ln bc

(a) (b)

From Eqs. (a) and (b), after eliminating pb , we obtain

pu = (σ yp ) i ln ab + (σ yp ) o ln bc 50 = 280 ln 30 20 + 400 ln 30 = 317.9 MPa

______________________________________________________________________________________ SOLUTION (12.49) ( a ) Equation (12.60):

pc = kσ yp 32 (−32)2 = 0.2778kσ yp 2

( b ) Equation (12.61a):

(σ r ) r =a = kσ yp [ln 12 − 0.2778] = −0.9709kσ yp

( c ) Equation (12.59b):

= (σ θ ) r =b Equation (12.59b):

= (σ θ ) r =c Equation (12.61b):

0.2778 kσ yp (22 ) 32 − 22

= (1 + 332 ) 0.444kσ yp

0.2778 kσ yp (22 ) 32 − 22

= (1 + 232 ) 0.7222kσ yp 2

(σ θ ) r =a = kσ yp (1 + ln 12 − 0.2778) = 0.0291kσ yp

We see from these results that the maximum stress occurs at the elastic-plastic boundary.

End of Chapter 12

CHAPTER 13 SOLUTION (13.1) x

b y ( a ) Boundary conditions at y=0 and y=b:

w=0

dw dy

(a)

We have 2

w' ' ' ' = pD0

w' ' = p20Dy + c1 y + c2

w' = p60Dy + 12 c1 y 2 + c2 y + c3 4

w = p240 yD + c16y + c22y + c3 y + c4 Conditions (a) yield c4 = 0, c3 = 0, 2

c1 = − 2p0Db Thus,

p0b c2 = 12 D

p0b y 4 y 3 y 2 w = 24 D [( b ) − 2( b ) + ( b ) ]

( b ) Differentiating twice the foregoing expression, we have 4

p0b 2 12 12 2 w' ' = 24 D [ b 4 y − b3 y + b 2 ]

For y = b2 :

d 2w dy 2

p0b = − 24 D 4

Hence,

0b M y = p24

Thus,

σ y ,max = p4 ( bt ) 2 , 0

σ x ,max = 12p ( bt ) 2 0

Similarly, for y=0: d 2w dy 2

and

p0b = − 12 D ,

M y = p120b

σ y ,max = 12 p0 ( bt ) 2 = σ max

σ x ,max = 16 p0 ( bt ) 2 ______________________________________________________________________________________ SOLUTION (13.2) We have rx = r ,

ry = ∞

and = = r 0.12 m, t 0.3(10−3 ) m Equation (13.3b):

ε max=

t 2r

= 1250 µ

0.3 2(120)

Equation (13.5): 9

×10 (0.3) σ max = − 2(1−Etν ) r = − 200 274.7 MPa 2(0.91)120 = 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.3) Using Eq. (13.8), 9

)( 0.012 ) D = 9 ( 200×10 = 32,400 12 ( 8 )

From Example 13.1:

w = ( πb ) 4 pD0 sin( πby ) −3 0.6 4 20×10 = wmax (= 0.82(10 = ) m 0.82 mm π ) 32,400 3

M y = − D ddyw2 = −( πb ) 2 p0 sin πby 2

Thus,

σ y ,max =

6 M y , max t2

= 0.61 p0 ( bt ) 2

0.6 2 = 0.61(20 ×103 )( 0.012 ) = 30.5 MPa

= σ x ,max ν= (30.5) 10.17 MPa Then,

ε y ,max = E1 (σ y ,max − νσ x ,max ) = 2001×103 (30.5 − 10.17 3 ) =136 µ

and

= ry

t 2ε y ,max

0.012×106 2(136)

= 44.42 m

______________________________________________________________________________________ SOLUTION (13.4) πy

πy

Dw' ' = −( πb ) 2 p0 sin b + c1 y + c2

Dw IV = p0 sin b

(a)

πy

Thus

Dw' = ( πb )3 p0 cos b + 12 c1 y 2 + c2 y + c3

Dw''' = −( πb ) p0 cos b + c1 πy

Dw = ( πb ) 4 sin b + 16 c1 y 3 + 12 c2 y 2 + c3 y + c4 Boundary conditions:

w' ( 0) = 0:

c3 = −( πb )3 p0 ;

w( b) = 0:

c2 = − c1b + π 3 p0 b

w' ( b) = 0:

c1 = 0

1 3

c4 = 0

w( 0) = 0;

(a)

∴ c2 = π23 p0 b 2

Equation (a) gives then p b4

πy

w = D0π 2 [sin b + bπ2 y 2 − πb y ] ( b ) For y= − b 2 : At y= 0:

(b)

p b4

wmax = π04 D (1 − π4 ) 2

ε max = z ddyw = 2t [0 + 0 + 2πp Db ] = pπ tbD 2

0 3

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.5)

Dw IV = p0,

(a)

Dw' ' = 12 p0 y 2 + c1 y + c2

Dw' ' ' = p0 y + c1 ,

Dw' = 16 p0 y 3 + 12 c1 y 2 + c2 y + c3

and

Dw = 241 p0 y 4 + 16 c1 y 3 + 12 c2 y 2 + c3 y + c4

Boundary conditions:

w( 0) = 0: c4 = 0,

(a)

w' ' ( 0) = 0: c2 = 0

w( b) = 0:

p0 b4 24

+ b6 c1 + c3b = 0

w' ( b) = 0:

p0 b3 6

+ b2 c1 + c3 = 0

Solving, c1 = − 83 b Equation (a) is thus

b c3 = 48

p b4

4 y3

9 y2

w = 480 D [( b ) − 3( b )3 + 2( b ) 4 ]

(b)

( b ) We have dw dy

= 240 D [ b4 − 2 b3 + 21b ],

d 2w dy 2

At y = b: 2

M max = − D ddyw2 = − p0 b8 ,

3 yb

= 2 D0 [ y 2 − 4 ]

σ y ,max = 6 Mt

max 2

= −0.75 p0 ( bt )2

______________________________________________________________________________________ SOLUTION (13.6) p

Dw IV = b0 y , Dw' ' ' = 2 b0 y 2 + c1 ,

(a)

Dw' ' = 6 b0 y 3 + c1 y + c2

Dw' = 240b y 4 + 12 c1 y 2 + c2 y + c3 p

Dw = 1200b y 5 + 16 c1 y 3 + 12 c2 y 2 + c3 y + c4 Boundary conditions:

w( 0) = 0: c4 = 0, p0 b3 24

w' ' ( 0) = 0: c2 = 0 2

+ b2 c1 + c3 = 0 , 1 1 Solving, c1 = − 10 p0 b c3 = 120 p0 b 3 w' ( b) = 0:

Thus

(a)

w( b) = 0:

p0 b4 120

+ b6 c1 + c3b = 0

w = 1200Db [ y 5 − 2b 2 y 3 + b 4 y ] ( b ) At

y = 0:

( c ) At

y = b:

(b)

p0 b3

θ = w' ( 0) = 120 D 2

(c)

M max = D ddyw2 = 60 b 2 − 101 p0 b 2 = 151 p0 b 2 and

σ y ,max = 6 Mt

max 2

3 2 2 = 52 p0 ( bt ) 2 = 80 MPa 5 (80 × 10 )(50) =

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.7)

M y = Ma

M x = Mb y

a ( a ) Using Eq. (13.7), ∂2w ∂x 2 2

∂ w ∂y 2

= − MD (b1−−ννM2 a) M a −νM b D (1−ν 2 )

=−

(a)

∂ w ∂x∂y

Integrating these equations,

w = − 2MDb(−1ν−Mν 2a) x 2 − 2MDa(−1ν−Mν 2b) y 2 + c1 x + c2 y + c3

If the origin of xyz is located at the center and midplane of the plate, the c’s will vanish, and

w = − 2MDb(−1ν−Mν 2a) x 2 − 2MDa(−1ν−Mν 2b) y 2

(b)

( b ) By setting M a = − M b in Eq. (a): ∂2w ∂x 2

= − ∂∂yw2 = D (M1−aν ) = r1x = − r1y 2

Integrating and locating the origin of xyz, as in item (a):

w = 2 DM(1a−ν ) ( x 2 − y 2 ) ______________________________________________________________________________________ SOLUTION (13.8) Equation (13.19) becomes, a

P [ (b − y ) sin nπby dy ]( a − x ) sin maπx dx pmn = 144 a 4b 4 ∫ ∫

Integrating by parts,

P b a P pmn = 144 = mn144 π 2 a 2b2 a 4 b 4 nπ mπ

(a)

Substituting Eqs. (13.18) into ∇ w = p D :

a mn = π 4 D ( m2 pamn2 + n 2 b2 )2

(b)

Inserting Eqs. (a) and (b) into Eq. (13.18b), we find the required expression for the deflection. ______________________________________________________________________________________ SOLUTION (13.9) ( a ) From Eq. (13.19), we obtain

pmn = p0

Then, Eq. (13.20) becomes for a square plate (a=b): nπy

w = πp04aD ∑∑ ( ma2 + n 2 )b 4

sin mπx sin

(a)

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 13.9 (CONT.) At x = y = a 2 , m + n −1 2

wmax = πp04aD ∑∑ (( m−12)+ n 2 )2 4

(b)

) D = 21012×10(1−(00.3.025 = 300,480.77 2 )

Equation (a) is then, 4

p0 ( 3) 1 8(10 −3 ) = π 4 ( 300 [ 1 − 100 ] , 480.77 ) 4

p0 = 12.05 kPa

______________________________________________________________________________________ SOLUTION (13.10) The flexural rigidity of the plate is

D = 12 (Et1−ν 2 ) = 12200(1×−100.09t ) = 18.315 × 10 9 t 3 3

9 3

The maximum deflection occurs at the center of the plate. Equation (13.27) is thus 4

p0 a wmax = 64 D ;

(10 )( 0.05 ) 1.5 × 10 −3 = 6410(18 .315×109 t 3 )

−3 Solving, t 3.29(10 = = ) m 3.29 mm

______________________________________________________________________________________ SOLUTION (13.11)

Since Q = 0, Eq. (13.24c) becomes d dr

[ 1r drd ( r dw dr )] = 0

from which, after integration, 2

w = − c14r − c2 ln ar + c3

(a)

M r = D[ c21 − cr 22 + ν ( c21 + cr 22 )]

(b)

Substituting this into Eq. (13.24a): Boundary conditions

( M r ) r =a = M 0

w( 0 ) = 0

( dw dr ) r =0 = 0

yield c2 = 0 and

c1 = D2(1M+ν0 )

c3 = 2 DM(01a+ν )

Equation (a) becomes then,

w = 2 DM(10+ν ) ( a 2 − r 2 ) Introducing this into Eqs. (13.24a) and (13.24b) yields M r = M θ = M 0 . Hence,

σ r ,max = σ θ ,max = 6 tM 2

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.12) We have Qr = 0 and Eq. (13.24c) after integration gives 2

w = − c14r − c2 ln ar + c3

(a)

M r = D[ c21 − cr 22 + ν ( c21 + cr 22 )

(b)

Introducing this into Eq. (13.24a):

Boundary conditions

w( a ) = 0

M r ( b) = M 0

and Eqs. (a) and (b) result in 2

2 2

c1 = (1+ν2)(b bM2 −0 a 2 ) ,

c2 = (1−νa )(bbM2 −0a 2 )

2 2

c3 = 2 (1+νa ) bD (Ma 20 −b2 )

Carrying these into Eq. (a), we have the equation for deflection. ______________________________________________________________________________________ SOLUTION (13.13) From Example 13.3:

σ 1 = σ r ,max = − 43 p0 ( at ) 2 ,

σ 2 = − 14 p0 ( at ) 2 ,

σ3 = 0

Maximum shearing stress is then

τ max = 12 (σ 1 − 0) = − 83 p0 ( at ) 2

According to maximum shear stress theory: σ yp n

= 43 p0 ( at ) 2

Introducing the given data, 100 (106 ) n

= (0.4 × 10 6 )( 00..022 ) 2 ( 43 ) n = 3.33

______________________________________________________________________________________ SOLUTION (13.14) po 2a Solution proceeds as in Example 13.3. Boundary conditions are We have

( w) r = a = 0 D po

( M r ) r =a = 0

(a)

r w =64 − c24r + c4 4

(b)

Carrying Eqs. (a) and (b) into Eq. (13.24a), we obtain two equations. From these c2 and c4 are evaluated. In so doing, and substituting the values obtained into Eq. (b): po ( a 2 − r 2 ) 64 D

( 15++νν a 2 − r 2 ) Center point deflection occurs at r = 0 .

= w

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.15) then rx = rxy = ∞ ,

Let r = ry ,

M x = M xy = 0

M = My,

Hence, Eq.(13.5), for z= − t 2 : 9

)(0.006) , 120(106 ) = 70(10 2(0.91) r

σ max = 2(1−Etν )r ; 2

Equation (13.9),

6 M max

6 120(10 = )

6 M max

= t2

(0.006)2

r=1.923 m, d= 3.85 m

M max = 720 N

;

______________________________________________________________________________________ SOLUTION (13.16) y

Let W = 1 − ax − a .

a a

w = cx 2 y 2W . 2 2 2 2 ∂w 1 ∂x = 2cxy W + 2cx y W ( − a )

Then

y=a−x

y = 4cxyW + 4cxy 2W ( − a1 ) + 4cx 2 y ( − a1 ) + 2cx 2 y 2W −1 ( − a1 )

∂ 2w ∂x∂y ∂w ∂y

= 2cx 2 yW + 2cx 2 y 2W ( − a1 )

∂ 2w ∂x 2

= 2cy 2W 2 + 4cxy 2W ( − a1 ) + 4cxy 2 ( − a1 ) + 2cx 2 y 2W −1 ( − a1 ) 2

∂ 2w ∂y 2

= 2cx 2W 2 + 4cx 2 yW ( − a1 ) + 4cx 2 y ( − a1 ) + 4cx 2 y 2W −1 ( − a1 ) 2

( a ) At x=0:

w=0,

At y=0:

w=0,

At y=a-x: w=0,

=0 =0

∂w ∂x ∂w ∂y ∂w ∂x

= 0, ∂∂wy = 0

( b ) At x=0, y=a: 2

σ y = − 2(1Et−ν ) [ ∂∂yw + ν ∂∂xw ] = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 0 , 2

∂ 2w ∂y 2

x = a2:

At y=0, and

∂ 2w ∂x 2

= ca8 , 2

, σ y ,max = − 2(1Et−ν ) [ ca8 ] = − 16Ecta (1−ν ) 2

∂ 2w ∂x∂y

= 0,

τ xy = 2(1E+ν ) ∂∂x∂wy = ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ = 0 =0

τ xy = 0

______________________________________________________________________________________ SOLUTION (13.17) ∂2

We have M xy = − D(1 − ν ) ∂x∂wy = M 0 Let,

∂ 2w ∂x∂y

−M

= D (1−ν0 ) = k

Integrating with respect to x : ∂w ∂y

= kx + f ( y ) + c1

Then, integrating the above with respect to y gives

w = kxy + ∫ f1 ( y )dy + c2 Due to the symmetry in deflection :

where

f1 ( y ) = f ( y ) + c1

∫ f ( y )dy = 0. 1

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 13.17 (CONT.) Also, owing to the symmetry, center (a/2, a/2) should be free of displacement,

∴ c2 = − 14 ka 2

w = 0 = 14 ka 2 + c2

It follows that

w = − D (1−0ν ) ( xy − a4 ) We observe that this solution satisfies boundary conditions, M x = 0 and M y = 0 at plate edges. ______________________________________________________________________________________ SOLUTION (13.18) Cylinder end can be approximated as a clamped edge plate subjected to uniform loading. ( a ) Equation (13.29), 3 p0

σ r ,max = 4 ( at ) 2 ;

135(106 )×4

= ( at ) 2

= 120

3×1.5(106 )

or a/t=10.954. Hence,

= t

= 18.26 mm

200 10.954

( b ) Then, Eq.(13.27) for r=0, gives p0 a 4 64 D

wmax =

1.5×106 (0.2)4 ×12(1− 0.32 )

= 0.336 mm

64×200×109 (0.01826)3

______________________________________________________________________________________ SOLUTION (13.19) ( a ) From Eq. (13.29):

= σ max

= p0 ( at ) 2

3 4

125 2 p= 117.19 p yp yp ( 10 )

Setting σ max = σ yp σ

= p yp We have = D

yp =

117.19

Et 3 12(1−ν 2 )

= 2.944 MPa

345 117.19

200(109 )(0.01)3

= 18.315 kPa

12(1− 0.32 )

Eq. (13.27) for r=0 is then

w= max (b)

p yp n

pallow =

p yp a 4 64 D

2.944(106 )(0.125)4

= 0.613 mm

64(18.315×103 )

= 2.45 MPa

2.944 1.2

______________________________________________________________________________________ SOLUTION (13.20) Referring to Example 13.2:

σ max = 6 Mt

max 2

= 6(0.0534 p0 )( 502 ) 2

Thus,

240(10 6 ) = 200.25 p0

p0 = 1.2 MPa

Similarly,

wmax = 0.0454 p0 Eta 3 = 0.0454(1.2 × 10 6 ) 70×10( 09.5( 0) .02 )3 4

= 0.0608 = m 6.1 mm ______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.21) We observe that a

∫ sin π sin π dx = ∫ sin m x a

m x a

nπy b

sin nπby dy = 0

if m ≠ m' and n ≠ n ' Therefore, integrating, we consider only the squares of the terms in the parenthesis in Eq. (b) of Sec. 13.9. Using the formula: a

∫ ∫ sin

2 mπx a

sin 2 nπby dxdy = ab4

calculation of the first term of the integral in Eq. (b) gives π 4 abD 8

∞

∑∑ a ( 2 mn

+ nb2 ) 2

m2 a2

Also, the second term of the integral in Eq. (b): a

p0 ∫ ∫ a mn sin maπx sin nπby dxdy ab = πp20mn a mn (1 − cos mπ )(1 − cos nπ )

= 4π p20mnab a mn

( m, n = 1, 3, . . . )

______________________________________________________________________________________ SOLUTION (13.22) Deflection is given by Eq. (13.18b). Loading is expressed as follows:

p = 2 p0 ax p = 2 p0 − 2 p0 ax

Potential energy, Eq. (13.33):

W = 2∑∑ ∫

a 2

∫

0 ≤ x ≤ a2 a 2 ≤ x ≤ a

a 2p x 0

(a)

sin maπx sin naπy dxdy

= ∑∑ m8 p2n0aπ 3 a mn sin m2π 2

(b)

Strain energy, Eq. (13.32): a

U = D2 ∫ ∫ [a mn ( maπ2 + nbπ2 ) sin maπx sin nπay ]2 dxdy 2 2

2 2

2 ( ma 2 + an 2 ) 2 = 81 Dπ 4 a 2 ∑∑ a mn 2

We thus have ∂Π ∂amn

= Dπ4 a a mn ( ma 2 + an 2 ) 2 − m8 p2n0aπ 3 sin m2π = 0 4 2

mπ 2 ) a mn = m322npπ0a7 Dsin( ( m2 + n 2 )2

( m, n = 1, 3, . )

(c)

Substitute this into Eq. (13.18b) to obtain deflection. ______________________________________________________________________________________ SOLUTION (13.23) Let p represent the pressure differential. At crown, φ = 0 :

ap or σ= −σ φ = σθ = t

= σ

150 p 6

= 25 p

(CONT.) ______________________________________________________________________________________ 343 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 13.23 (CONT.) From the above, p = σ / 25 = 15 ×10 / 25 = 600 kPa 6

At edge, φ = 90 : o

pa σ= −σ θ = −σ φ = 12.5 p 2t =

Thus,

p= 15 ×106 12.5 = 1.2 MPa σ 12.5 =

______________________________________________________________________________________ SOLUTION (13.24) The circumferential, axial, and radial stresses are given by

σ= 1

pr t

= 30 p σ= 2

pr 2t

0 = 15 p σ= 3

Insertion of these expression into Eqs. (4.1) and (4.5a) provide the critical pressures. ( a ) For the maximum shearing stress theory:

30 p= − 0 1.51 (250 ×106 ) p = 5.56 MPa

( b ) For the maximum energy of distortion theory:

p (302 − 30 ×15 + 152 = )1 2 p = 6.42 MPa

1 1.5

(250 ×106 )

Comment: The permissible value of the internal pressure is conservatively limited to 5.56 MPa. ______________________________________________________________________________________ SOLUTION (13.25) We have σ = σ u

= t

pr 2σ u n

n . Applying Equation 13.48a, 15(0.8) 2(400 2.4)

= 0.036= m 36 mm .

Then Equation 13.48b result in

= δs

pr 2 (1−ν ) 2 Et

(15×106 )(0.8)2 (0.7)

−3 = 0.47(10= ) m 0.47 mm .

2(200×109 )(0.036)

The diametral extension is thus 2δ s = 0.94 mm ______________________________________________________________________________________ SOLUTION (13.26)

= p

2tσ all 2(0.05)(30 ×106 ) 2 = = 1.5 MPa 1 r

We have

9.81(103 )h = 1.5 ×106 ,

h= 152.9 m

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.27) y p r

Total pressure at any depth:

h x

= p 500(103 ) + γ (h − y ) (a) At y = h : p = 500(103 ) Pa

Liquid pressure Therefore,

t = (b)

pr 500(103 )4000 = = 11.11 mm 180(106 ) σ all

At y = h 4 :

p= 500(103 ) + 15(103 )(13.5) = 702.5(103 ) Pa

= t (c)

At y = 0 :

702.5(103 )4000 = 15.6 mm 180(106 )

p= 500(103 ) + 15(103 )(18) = 770(103 ) Pa

= t

770(103 )4000 = 17.1 mm 180(106 )

______________________________________________________________________________________ SOLUTION (13.28) Total pressure at any depth:

p 200(103 ) + γ (h − y ) = (a) At y = h : p = 200(103 ) Pa pr 200(103 )4000 t = = = 4.44 mm 180 ×106 σ all (b) At y = h 4 : p= 200(103 ) + 15(103 )(13.5) = 402.5(103 ) Pa = t (c)

At y = 0 :

402.5(103 )4000 = 8.94 mm 180 ×106

p =200(103 ) + 15(103 )18 =470(103 ) Pa = t

470(103 )4000 = 10.4 mm 180 ×106

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.29) The thickness for circumferential stress:

pr 1.2(103 )(0.6 ×103 ) = = 7.2 mm 240(106 ) 2.4 σ all

= t

The thickness for axial stress:

pr 7.2 = = 3.6 mm 2σ all 2 treq = 3.6 mm

t = Thus,

______________________________________________________________________________________ SOLUTION (13.30)

pr p γh = t pr γ hr 9.81(103 )(150)(400) treq = = = = 8.83 mm σ all σ all 120(106 ) 1.8

= σθ

______________________________________________________________________________________ SOLUTION (13.31)

treq =

pr γ hr 9.81(103 )(150)(400) = = = 10.6 mm σ all σ all 100(106 ) 1.8

______________________________________________________________________________________ SOLUTION (13.32) (a)

σ −σ 1 pr pr pr pd τ max = 1 2 = ( − ) = =

(b)

σ= 1

2 t 2t 4t 8t 6 8tτ max 8 ×12 × 35 ×10 d = = = 336 mm p 10 ×106

pr 10 ×106 ×168 = = 140 MPa t 12

______________________________________________________________________________________ SOLUTION (13.33)

F Nφ The given numerical values are:

rφ= 180 − 1= 179 mm

rθ = 80 − 1 = 79 mm p = −0.08 MPa

ri = 80 − 2 = 78 F = πri 2 p

(CONT.) ______________________________________________________________________________________ 346 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 13.33 (CONT.) Equation (13.46b) is therefore 2

) ( 0.08×10 ) N φ = 2πrFθ (1) = π ( 0.078 2π ( 0.079 )

and

= 3.081 kN m 3081 σ φ 0.002 = = 1.541 MPa

Using Eq. (13.46a), σθ

0.079

0.08 + 1.541 0.179 = 0.002

= σ θ 2.48 = MPa σ max

______________________________________________________________________________________ SOLUTION (13.34) b

Nφ

A a

Nθ

rθ r0

Consider the portion of shell defined by φ . Vertical equilibrium of force yields,

2πr0 N φ sin φ = πp( r02 − b 2 ) from which 2

pa ( r0 + b ) −b ) N φ = p2(rr00sin 2 r0 φ =

N φ = b+ apasin φ ( a2 sin φ + b) Substituting N φ into Eq. (13.46a), setting p z = − p, and rφ = a :

N θ = prθ 2( rr00−b ) = pa2 Since sin φ = r0

rθ = ( r0 − b) a , from symmetry.

Note that N θ is constant throughout the shell from the condition of symmetry. ______________________________________________________________________________________ SOLUTION (13.35) Referring to Solution of Prob. 13.34, we have

2a = (1 − 0.7) 2, 2b = (1 + 0.7) 2,

a= 0.075 m b= 0.425 m

At point A (crotch):

σ A = σ φ ,max = pa ( r0 + b) ( 2r0 t )

(CONT.) ______________________________________________________________________________________

______________________________________________________________________________________ 13.35 (CONT.) or

t = pa ( r0 + b) ( 2r0σ φ ,max ) 6

.075 )( 0.35+ 0.425 ) = 2 (102)(( 00.35 )( 210×106 )

= 0.791 = mm treq. Similarly,

σ θ = pa 2t

or = t

2(10 )75 = 0.357 mm

2(210×106 )

______________________________________________________________________________________ SOLUTION (13.36) The stresses in the brass and steel tubes are tensile and compressive, respectively, as α b > α s and ∆T is a negative quantity because of cooling. Substituting the given numerical values into Eq.(13.61), we obtain the stress in brass tube, we have 9

−6

(140)(18.9 −11.7)(10 ) (σ θ )b = − 1031×+10(0.005 = − 103.824 −55.9 MPa 1.8583 = ×103 0.003×200)

Similarly, Eq.(13.62) gives 9

−6

−18.9)(10 ) (σ θ ) s = − 200×10 (140)(11.7 = 108.5 MPa 1.8583

as the hoop stress in the steel tube. ______________________________________________________________________________________ SOLUTION (13.37) Nφ

Nφ

v φ

r2 α α

t y

O Pressure at any level st is p r = γ ( a − y ) . We have

φ = π2 + α

r0 = y tan α

Thus, the first of Eqs. (13.49) becomes and

y ) y tan α N θ = γ ( a −cos α

tan α σ θ = γ ( a−t y ) y cos α

The load F is equal to the weight of liquid in cylindrical portion stuv:

F = −πγy 2 ( a − y + 3y ) tan 2 α

Then second of Eqs. (13.49) gives

N φ = γy ( a − 23 y ) tan α 2 cos α

and

tan α σ φ = γy ( a −22t y 3) cos α

______________________________________________________________________________________

______________________________________________________________________________________ SOLUTION (13.38) Expressions for the components of pressure are:

pθ = − p cos θ

pr = p sin θ

Thus,

px = 0

N xθ = − ∫ [ − p cos θ + a1 ( − pa cos θ )]dx + f1 (θ ) = 2 px cos θ + f1 (θ ) N θ = − pa sin θ N x = − ∫ a2 ( − px sin θ )dx +

(a) (b) df1 dθ

+ f 2 (θ )

= a1 px 2 sin θ + dfdθ1 + f 2 (θ ) Boundary conditions: N x = 0 2

pL a

(c)

(at x=0 and x=L)

give f 2 = 0 and

sin θ + L dfdθ1 = 0

f1 (θ ) = − pL2 sin θ + c Note that no torque is applied to the shell; c = 0. Hence,

N θ = − pa sin θ N x = − La− x px sin θ N xθ = −( L − x ) p cos θ ______________________________________________________________________________________ SOLUTION (13.39) Now the cylinder length does not change:

∫

−L 2

( N x − νN θ )dx = 0

Substituting Eqs. (b) of Example 13.8 into this, taking f 1 = 0 and integrating the resulting expression, we have 2

f 2 (θ ) = νγa 2 (1 − cos θ ) − γ24L cos θ Referring to Example 13.8, the solution is thus,

N xθ = γax sin θ

N θ = γa 2 (1 − cos θ ) 2

N x = γx2 cos θ + νγa 2 (1 − cos θ ) − γ24L cos θ

______________________________________________________________________________________ SOLUTION (13.40) Referring to Fig. 13.15b:

p x = p sin φ

p z = p cos φ

r0 = x cos φ

Stress resultants due to weight:

F = 2πr ⋅ p sin φ ⋅ rdφ

(CONT.) ______________________________________________________________________________________ 349 From Advanced Mechanics of Materials and Applied Elasticity, 6/e, by Ansel C. Ugural and Saul K. Fenster (9780135793886) Copyright © 2020 Pearson Education, Inc. All rights reserved.

______________________________________________________________________________________ 13.40 (CONT.) Since

rdφ = dx and r = x cot φ . Then x

F = 2π ∫ x cot φp sin φdx = 2πp cos φ ( x2 ) + c 2

For a cone supported at ts edge c = 0 , since F = 0 at x = 0. Therefore, Eq. (13.46b) gives px N φ = − 2 sin φ

(1)

Equation (13.46a): 2

p z r0 cos φ N θ = sin φ = − px sin φ

(2)

Stress resultants due to pressure: Equation (13.46a) yields,

cos φ N θ = − p r xsin φ = − p r x cot φ

(3)

We now have

F = ( 2πp r sin φrdφ ) cos φ

Following a procedure similar to that the preceding, we obtain

F = 2πp r cos 2 φ ( x2 ) 2

Equation (13.46a) leads to

N φ = − p2r xsincosφ φ = − 12 p r x cot φ

(4)

Solution is determined by the superposition of the preceding results: adding Eqs. (4) and (1), and (3) and (2). In so doing, we have

N x = − 2 sinx φ ( p + p r cos φ )

N θ = − x cot φ ( p cos φ + 12 p r )

(P13.40)

End of Chapter 13

712

MATLAB SOLUTIONS of Problems Listed in Table E.1 (page 712 in text) by

Ansel C. Ugural and Youngjin Chung

%EXAMPLE 1.1 State of Stress in a Tensile Bar %Given: The stresses on the inclined plane with theta = 35o for a prismatic bar of a cross-sectional area %800 mm^2, subjected to a tensile load of 60 kN (Fig. 1.6a). theta=35*pi/180.; A=800*10^-6; P=60*10^3; %Find: The state of stress for by calculating the stresses on an adjoining face of a stress element. Sketch %the stress configuration. %Solution The normal stress on a cross section is sigma_x=P/A %Introducing this value in Eqs. (1.11) and using we have sigma_x_p=sigma_x*(cos(theta))^2 tau_x_p_y_p=-sigma_x*sin(theta)*cos(theta) %Comments: The normal and shearing stresses acting on the adjoining y' face are, respectively, 24.67 %MPa and 35.24 MPa, as calculated from Eqs. (1.11) by substituting the angle theta + 90o = 125o The %values of sigma_x' and tau_x'y' are the same on opposite sides of the element. On the basis of the %established sign convention for stress, the required sketch is shown in Fig. 1.8. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 1.1: sigma_x = 75000000 sigma_x_p = 5.0326e+07 tau_x_p_y_p = -3.5238e+07 Results in textbook notation with units:

 x ' =50.33 MPa  x ' y ' =-35.24 MPa >> %EXAMPLE 1.5 Cylindrical Vessel Under Combined Loads %Given: A thin-walled cylindrical pressure vessel of 250-mm diameter and 5-mm wall thickness rigidly %attached to a wall, forming a cantilever (Fig. 1.18a). d=0.25; r=d/2.; t=0.005; p=1.2*10^6; T=3*10^3; P=20*10^3; %Find: the maximum shearing stresses and the associated normal stresses at point A of the cylindrical %wall. The following loads are applied: internal pressure p = 1.2 MPa, torque T = 3 kN∙m and direct force %P = 20 kN. Show the results on a properly oriented element. %Solution

%The internal force resultants on a transverse section through point A are found from the equilibrium %conditions of the free-body diagram of Fig. 1.18b. They are V = 20 kN, M = 8 kN∙m, and T = 3 kN∙m. In %Fig.1.18c, the combined axial, tangential, and shearing stresses are shown acting on a small element at %point A. These stresses are (Tables 1.1 and C.1) V=20*10^3; M=8*10^3; I=pi*r^3*t; J=2*pi*r^3*t; sigma_b=M*r/I tau_t=-T*r/J sigma_a=p*r/(2*t) sigma_theta=2*sigma_a %We thus have Q=0; tau_d=V*Q/(I*t); tau_xz=V*Q/(I*t); sigma_x=sigma_a+sigma_b sigma_y=sigma_theta tau_xy=tau_t %The maximum shearing stresses are from Eq. (1.22): tau_max=sqrt(((sigma_x-sigma_y)/2)^2+tau_xy^2) tau_max_m=-tau_max %Equation (1.23) yields sigma_p=(sigma_x+sigma_y)/2 %To locate the maximum shear planes, we use Eq. (1.21): theta_s=(1/2.)*atan(-(sigma_x-sigma_y)/(2*tau_xy)); theta_s_d=theta_s*180/pi theta_s_add=theta_s_d+90. %Applying Eq. (1.18b) with the given data tau_x_p_y_p=-(1/2.)*(sigma_x-sigma_y)*sin(2*theta_s)+tau_xy*cos(2*theta_s) theta_pp=theta_s_d ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 1.5: sigma_b = 3.2595e+07 tau_t = -6.1115e+06 sigma_a = 15000000 sigma_theta = 30000000 sigma_x = 3

4.7595e+07 sigma_y = 30000000 tau_xy = -6.1115e+06 tau_max = 1.0712e+07 tau_max_m = -1.0712e+07 sigma_p = 3.8797e+07 theta_s_d = 27.6063 theta_s_add = 117.6063 tau_x_p_y_p = -1.0712e+07 theta_pp = 27.6063 Results in textbook notation with units:

 b = 32.6 MPa,  t = −6.112 MPa,  a = 15 MPa,  a = 30 MPa,  max = 10.71 MPa,  ' = 38.8 MPa, s = 27.6o and 117.6o ,  x ' y ' = −10.71 MPa >> %CASE STUDY 1.1 Pressure Capacity of an Hydraulic Cylinder %Pressurized hydraulic fluid (liquid or air) produces stresses and deformation of a cylinder. Hydraulic %systems are widely used in brakes, control mechanisms, and actuators in positioning devices. Design of %a pressurized duplex conduit is discussed in Example 8.4. %Given: A hydraulic cylinder of radius r and thickness t under internal pressure p is simultaneously %compressed by an axial load P through the piston of diameter d≈2r (Fig. 1.19a). Observe that, the %vessel is inadvertently subjected to torque T at its mounting. Data: r = 60 mm, t = 4 mm, and T = 260 %N·m. 4

tau_yp=260*10^6; r=0.06; t=0.004; T=260; n=2; %Find: The largest value of p that can be applied to the cylinder with a safety factor of n. Assumptions: %The vessel is made of a high strength ASTM-A242 steel with tau_yp = 260 MPa (Table D.1). The critical %stress is at point A on cylinder wall remote from the ends (Fig. 1.19b). The effect of bending of the %cylinder on stresses is omitted. %Solution: Combined stresses acting at point A on an element in this thin-walled cylinder are found as J=2*pi*r^3*t; tau_xy=T*r/J sigma_x_div_p=r/(2*t) sigma_th_div_p=r/t %Maximum allowable in-plane shear stress in the cylinder wall will be 210/2 =105 MPa. Through the use %of Eq. 1.24, we obtain %tau_max=sqrt(((sigma_x_div_p-sigma_yh_div_p)/2)^2+tau_xy^2) tau_max=105*10^6; %Solving, p_max=sqrt( (tau_max^2-tau_xy^2)/(((sigma_x_div_p-sigma_th_div_p)/2)^2) ) %Comments: The largest permissible axial load that can be applied to piston is about P_max = 28(pi × %60^2) = 317kN. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Case Study 1.1: tau_xy = 2.8736e+06 sigma_x_div_p = 7.5000 sigma_th_div_p = 15 p_max = 2.7990e+07 Results in textbook notation with units:

 xy = 2.8736 MPa,  x = 7.5 p,   = 15 p, pmax = 28 MPa >> %EXAMPLE 1.6 Three-Dimensional Stress in a Hub %A steel shaft is to be force fitted into a fixed-ended cast-iron hub. The shaft is subjected to a bending %moment M, a torque T, and a vertical force P, Fig. 1.21a. Suppose that at a point Q in the hub, the %stress field is as shown in Fig. 1.21b, represented by the matrix 5

%Given: %[-19 -4.7 6.45 % -4.7 4.6 11.8 % 6.45 11.8 -8.3] MPa %Find: The principal stresses and their orientation with respect to the original coordinate system. %Solution %Substituting the given stresses into Eq. (1.33) we obtain from Eqs. (B.2) Sx=-19*10^6; Sy=4.6*10^6; Sz=-8.3*10^6; Txy=-4.7*10^6; Txz=6.45*10^6; Tyz=11.8*10^6; I1=Sx+Sy+Sz; I2=Sx*Sy+Sx*Sz+Sy*Sz-((Txy^2)+(Tyz^2)+(Txz^2)); I3=Sx*Sy*Sz+2*Txy*Tyz*Txz-Sx*(Tyz^2)-Sy*(Txz^2)-Sz*(Txy^2); R=(I1^2)/3.-I2; T=sqrt((R^3)/27.); Q=I1*I2/3.-I3-(2/27.)*(I1^3); ST=sqrt(R/3.); alpha=acos(-Q/(2.*T)); S(1)=2*ST*cos(alpha/3.)+I1/3.; S(2)=2*ST*cos((alpha/3.)+120*pi/180.)+I1/3.; S(3)=2*ST*cos((alpha/3.)+240*pi/180.)+I1/3.; for i=[1,2] for j=[i,3] if(S(i) <= S(j)) temp=S(i); S(i)=S(j); S(j)=temp; end end end sigma_1=S(1) sigma_2=S(2) sigma_3=S(3) %Successive introduction of these values into Eq. (1.31), together with Eq. (1.30a), or application of Eqs. %(B.6) yields the direction cosines that define the orientation of the planes on which sigma_1, sigma_2, %and sigma_3 act: for i=[1,3] a(i)=((Sy-S(i))*(Sz-S(i))-(Tyz^2)); b(i)=-(Txy*(Sz-S(i))-(Txz*Tyz)); c(i)=((Txy*Tyz)-Txz*(Sy-S(i))); k(i)=1./(sqrt((a(i)^2)+(b(i)^2)+(c(i)^2))); l(i)=a(i)*k(i); m(i)=b(i)*k(i); n(i)=c(i)*k(i); end l1=l(1) 6

m1=m(1) n1=n(1) l2=l(2) m2=m(2) n2=n(2) l3=l(3) m3=m(3) n3=n(3) %Note that the directions of the principal stresses are seldom required for purposes of predicting the %behavior of structural members. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 1.6: sigma_1 = 1.1618e+07 sigma_2 = -9.0015e+06 sigma_3 = -2.5316e+07 l1 = 0.0266 m1 = -0.8638 n1 = -0.5031 l2 = -0.6209 m2 = 0.3802 n2 = -0.6855 l3 = 7

0.7834 m3 = 0.3306 n3 = -0.5262 Results in textbook notation with units:

 1 = 11.618 MPa,  2 = −9.002 MPa,  3 = −25.316 MPa l1 = 0.0266, l2 = −0.6209, l3 = 0.7834 m1 = −0.8638, m2 = 0.3802, m3 = 0.3306 n1 = −0.5031, n2 = −0.6855, n3 = −0.5262 >> %EXAMPLE 2.2 Three-Dimensional Strain in a Block %Given: A 2-m by 1.5-m by 1-m parallelepiped is deformed by movement of corner point A to A’ %(1.9985, 1.4988, 1.0009), as shown by the dashed lines in Fig. 2.7. %Find: The following quantities at point A: (a) the strain components; (b) the normal strain in the %direction of line AB; and (c) the shearing strain for perpendicular lines AB and AC. %Solution The components of displacement of point A are given by u_A=-1.5/1000.; v_A=-1.2/1000.; w_A=0.9/1000.; %a. We can readily obtain the strain components, by using an approximate version of Eqs. (2.4) x=2.; y=1.5; z=1.; c1=u_A/(x*y*z); c2=v_A/(x*y*z); c3=w_A/(x*y*z); c1_mu=c1*10^6; c2_mu=c2*10^6; c3_mu=c3*10^6; epsilon_x=c1_mu*y*z epsilon_y=c2_mu*x*z epsilon_z=c3_mu*x*y gamma_xy=c2_mu*y*z+c1_mu*x*z gamma_yz=c3_mu*x*z+c2_mu*x*y gamma_xz=c3_mu*y*z+c1_mu*x*y %b. Let the x’ axis be placed along the line from A to B. The direction cosines of AB are l1=-0.8; m1=-0.6; n1=0.; %and Applying Eq.(2.18a), we thus have 8

epsilon_x_p=epsilon_x*(l1^2)+epsilon_y*(m1^2)+gamma_xy*l1*m1 %c. Let the y’ axis be placed along the line A to C. The direction cosines. %Thus, from Eq. (2.18b), l2=0.; m2=0.; n2=-1.; gamma_x_p_y_p=gamma_yz*m1*n2+gamma_xz*l1*n2 %where the negative sign indicates that angle BAC has increased ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 2.2: epsilon_x = -750 epsilon_y = -800.0000 epsilon_z = 900 gamma_xy = -1600 gamma_yz = -600.0000 gamma_xz = -1050 epsilon_x_p = -1536 gamma_x_p_y_p = -1.2000e+03 Results in textbook notation with units:

(a)  x = −750 ,  y = −800 ,  z = 900 ,  xy = −1600 ,  yz = −600 ,  xz = −1050  (b)  x ' = −1536 , (c)  x ' y ' = −1200  >> %EXAMPLE 2.4 Deformation of a Tension Bar

%An aluminum alloy bar of circular cross-sectional area A and length L is subjected to an axial tensile %force P (Fig. 2.18). %Find: (a) the change in diameter delta_d ; (b) the change in volume delta_V . (c) Evaluate the numerical %values of the axial deformation delta, delta_d, and delta_V for the case in which P=60 kN, d=25 %mm, L=3 m, E=70 GPa, nu=0.3 and sigma_yp=260 MPa. %Given: P=60*10^3; d=0.025; L=3; E=70*10^9; nu=0.3; sigma_yp=260*10^6; A=pi*d^2/4 %Solution %a. The change in diameter delta_d=-nu*P*d/(A*E); %b. The change in volume delta_V=P*L*(1-2*nu)/E; %c.the axial stress sigma in the bar is obtained from sigma=P/A %which is well below the yield strength of 260 MPa. Thus, introducing the given data into the preceding %equations, we have delta=P*L/(A*E) delta_d=-nu*P*d/(A*E) delta_V=P*L*(1-2*nu)/E %Comment: A positive sign indicates an increase in length and volume; the negative sign means that the %diameter has decreased. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 2.4: A= 4.9087e-04 sigma = 1.2223e+08 delta = 0.0052 delta_d = -1.3096e-05 delta_V = 1.0286e-06

Results in textbook notation with units:

(c)  = 122.2 MPa,  = 5.2 mm, d = −0.0131 mm, V = 1029 mm 3 >> %EXAMPLE 2.8 Principal Strains on Surface of a Steel Frame %Strain rosette readings are made at a critical point on the free surface in a structural steel member. %The 60° rosette contains three wire gages positioned at 0°, 60° and 120° (Fig. 2.24b). The readings are % epsilon_a=190 mu, epsilon_b=200 mu, epsilon_c=-300 mu %Find: (a) The in-plane principal strains and stresses and their directions; (b) the true maximum shearing %strain. %Given: The material properties are E=200 GPa and nu=0.3. E=200*10^9; nu=0.3; %Solution For the situation described, Eq. (2.44) provides three simultaneous expressions: epsilon_a=190; epsilon_b=200; epsilon_c=-300; epsilon_x=epsilon_a epsilon_y=(2*(epsilon_b+epsilon_c)-epsilon_a)/3 gamma_xy=(2/sqrt(3))*(epsilon_b-epsilon_c) %a. Upon substituting numerical values, we obtain epsilon_x=190 mu, epsilon_y =-130 mu, and %gamma_xy=577 mu. Then, from Eq. (2.16), the principal strains are epsilon_1=((epsilon_x+epsilon_y)/2.)+sqrt(((epsilon_x-epsilon_y)/2.)^2+(gamma_xy/2)^2) epsilon_2=((epsilon_x+epsilon_y)/2.)-sqrt(((epsilon_x-epsilon_y)/2.)^2+(gamma_xy/2)^2) %The maximum shear strain is found from gamma_max1=(epsilon_1-epsilon_2) gamma_max2=-(epsilon_1-epsilon_2) %The orientations of the principal axes are given by Eq. (2.15): theta_p=atan(((gamma_xy)/(epsilon_x-epsilon_y))) theta_p=(theta_p*180./pi)/2. theta_p_p=theta_p+90. %Thus, the first two equations of (2.34) for plane stress sigma_1=(E/(1-nu^2))*(epsilon_1+nu*epsilon_2)*10^-6 sigma_2=(E/(1-nu^2))*(epsilon_2+nu*epsilon_1)*10^-6 %From Eq. (2.36), the maximum shear stress is G=E/(2*(1+nu)); tau_max=G*gamma_max1*10^-6 %b. Applying Eq. (3.11b), the out-of-plane principal strain is epsilon_z=-(nu/(1-nu))*(epsilon_x+epsilon_y) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 2.8: epsilon_x = 11

190 epsilon_y = -130 gamma_xy = 577.3503 epsilon_1 = 360.0505 epsilon_2 = -300.0505 gamma_max1 = 660.1010 gamma_max2 = -660.1010 theta_p = 30.5012 theta_p_p = 120.5012 sigma_1 = 5.9348e+07 sigma_2 = -4.2206e+07 tau_max = 5.0777e+07 epsilon_z = -25.7143 Results in textbook notation with units:

(a)  x = 190 ,  y = −130 ,  xy = 577 , 1 = 360 ,  2 = −300 ,  max = 660 

 p' = 30.5o ,  p" = 120.5o ,  1 = 59.35 MPa,  2 = −42.2 MPa,  max = 50.78 MPa (b)  z = −26  12

>> %EXAMPLE 3.5 Analysis of Cam and Follower %Given: A camshaft and follower of an intermittent motion mechanism is illustrated in Fig. 3.23. For the %position depicted, the cam exerts a force Fmax on the follower. %Find: (a) the maximum stress at the contact line between the cam and follower; (b) the deflection. %Given: F_max = 8 kN, rc = 40 mm, D_f = L = 35 mm, E = 200 GPa, and sigma_yp = 510 MPa. F_max=8000; r_c=0.04; D_f=0.035; L=D_f; E=200*10^9; sigma_yp=510*10^6; %Assumptions: The material of all parts is hardened on the surface. Frictional forces can be omitted. The %rotational speed is slow so that the loading is considered static. %Solution %Formulas on the second column of case A of Table 3.2 apply. We begin by calculating the half-width a %of the contact patch. Inasmuch as E1=E2=E and delta=2/E E1=E; E2=E1; delta=2/E; %hence, a=1.076*sqrt((F_max/L)*r_c*delta) %a. The largest contact pressure is therefore p_0=(2/pi)*(F_max/(a*L)) %b. The deflection delta of the cam and follower at the line of contact is given by delta=(0.579*F_max/(E*L))*(1/3.+log(2*r_c/a)) %Comments The maximum contact stress is calculated to be smaller than the yield strength of 510 %MPa; the design is satisfactory. Deflection obtained between the cam and the follower is very small %and does not affect the performance of the mechanism ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 3.5: a= 3.2535e-04 p_0 = 4.4725e+08 delta = 3.8632e-06 Results in textbook notation with units:

(a) a = 0.325 mm (b) po = 447 MPa,  = 0.0039 mm >> %EXAMPLE 3.6 Steel Railway Car Wheel 13

%A railway car wheel rolls on a rail. Both rail and wheel are made of steel for which and The wheel has %a radius of and the cross radius of the rail top surface is (Fig. 3.25b). %Determine the size of the contact area and the maximum contact pressure, given a compression load %of . E=210*10^9; nu=0.3; r1=0.4; r2=0.3; F=90*10^3; %Solution For the situation described, 1/r1=1/r2=0 and, because the axes of the members are mutually %perpendicular, theta=pi/2. The first of Eqs. (3.72) and Eqs. (3.74) reduce to m=4/(1/r1+1/r2) A=(1/r1+1/r2)/2.; B1=(1/r1-1/r2)/2.; B1=-(1/r1-1/r2)/2.; %The proper sign in B must be chosen so that its values are positive. Now Eq. (3.73) has the form n=4*E/(3*(1-nu^2)) cos_alpha=B1/A alpha=acos(cos_alpha)*180/pi %Corresponding to this value of alpha, interpolating in Table 3.3, we have c_a=1.1040; c_b=0.9113; %The semiaxes of the elliptical contact are found by applying Eqs. (3.71): a=c_a*(F*m/n)^(1/3) b=c_b*(F*m/n)^(1/3) p_0=1.5*(F/(pi*a*b)) %Comment A hardened steel material is capable of resisting this or somewhat higher stress levels for %the body geometries and loading conditions described in this section ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 3.6: m= 0.6857 n= 3.0769e+11 cos_alpha = 0.1429 alpha = 81.7868 a= 0.0065

b= 0.0053 p_0 = 1.2465e+09 Results in textbook notation with units:

m = 0.6857, n = 3.0769,  = 81.79o , a = 0.0065 mm, b = 0.0053 mm, po = 1247 MPa >> %EXAMPLE 4.3 Tube Torque Requirement %A thin-walled tube is fabricated of a brittle material having ultimate tensile and compressive strengths %sigma_u=300 MPa and sigma_u_p=700 MPa. The radius and thickness of the tube are r = 100 mm and t = 5 mm sigma_u=300*10^6; sigma_u_p=700*10^9; t=0.005; r=0.1; %Calculate the limiting torque that can be applied without causing failure by fracture. Apply (a) the %maximum principal stress theory and (b) the Coulomb[nd]Mohr theory. %Solution The torque and maximum shearing stress are related by the torsion formula J=2*pi*r^2*t; T_div_tau=J %The state of stress is described by sigma_1=-sigma_2=tau, sigma_3=0. %a. Maximum principal stress theory: %Equations(4.10) are applied with sigma_3 replaced by sigma_2 because the latter is negative: %|sigma_1|=|sigma_2|=sigma_u. Because we have sigma=sigma_u=300*10^6=tau from Eq. (f) tau=300*10^6; T=T_div_tau*tau; T_max_princ=T %b. Coulomb[nd]Mohr theory: Applying Eq. (4.12a), %from which tau = 210 MPa. Equation (f) gives tau=210*10^6; T=T_div_tau*tau; T_coulomb=T %Comment: Based on the maximum principal stress theory, the torque that can be applied to the tube %is thus 30% larger than that based on the Coulomb -]Mohr theory. To prevent fracture, the torque %should not exceed 65.9 kN·m. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 4.3: T_div_tau

3.1416e-04 T_max_princ = 9.4248e+04 T_coulomb = 6.5973e+04 Results in textbook notation with units:

(a) T = 94.2 kN  m, (b) T = 66 kN  m >> %EXAMPLE 4.4 Design of a Cast-Iron Torsion Bar %A torsion-bar spring made of ASTM grade A-48 cast iron is loaded as shown in Fig.4.11. The stress %concentration factors are 1.7 for bending and 1.4 for torsion. K_b=1.7; K_t=1.4; %Determine the diameter d to resist loads P = 25 N and T = 10 N·m, using a factor of safety n = 2.5. %Apply (a) the maximum principal stress theory and (b) the Coulomb[nd]Mohr theory. P=25.; T=10.; n=2.5; %Solution The stresses produced by bending moment M = 0.1P and torque T at the shoulder are M=0.1*P; %sigma_x=K_b*32*M/(pi*d^3) %tau_xy=K_t*16*T/(pi*d^3) %The principal stresses, using Eq. (4.7), are then sigma_1_div_d_1_3=(16/pi)*(K_b*M+sqrt((K_b*K_b)*(M*M)+(K_t*K_t)*(T*T))) sigma_2_div_d_2_3=(16/pi)*(K_b*M-sqrt((K_b*K_b)*(M*M)+(K_t*K_t)*(T*T))) %for system accepting minus value sigma_2_div_d_2_3=abs(sigma_2_div_d_2_3); %The allowable ultimate strengths of the material in tension and compression are 170/2.5 = 68 MPa and %650/2.5 = 260 MPa, respectively (see Table D.1). sigma_all_ten=170*10^6/n sigma_all_comp=650*10^6/n; %choose tension case %a. Maximum principal stress theory: On the basis of Eqs. (g) and (4.10), d_1_max_princ=((sigma_1_div_d_1_3)/sigma_all_ten)^(1/3) d_2_max_princ=((sigma_2_div_d_2_3)/sigma_all_ten)^(1/3) %b. Coulomb[nd]Mohr theory: Using Eqs. (g) and (4.12a), d_coulomb=((sigma_1_div_d_1_3/sigma_all_ten)+(sigma_2_div_d_2_3/sigma_all_comp))^(1/3) %Comments The diameter of the spring based on the Coulomb[nd]Mohr theory is therefore about 4.5% %larger than that based on the maximum principal stress theory. A 12-mm-diameter bar, a commercial %size, should be used to prevent fracture.

---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 4.4: sigma_1_div_d_1_3 = 96.1595 sigma_2_div_d_2_3 = -52.8694 sigma_all_ten = 68000000 d_1_max_princ = 0.0112 d_2_max_princ = 0.0092 d_coulomb = 0.0117 Results in textbook notation with units:

1 =

96.16 52.87 ,  2 = − 3 (a) d = 11.2 mm, d = 9.2 mm (b) d = 11.7 mm 3 d d

>> %EXAMPLE 4.6 Design of Titanium Panel with a Central Crack %A long plate of width 2w is subjected to a tensile force P in longitudinal direction with a safety factor of %n (see case A, Table 4.2). %Determine the thickness t required (a) to resist yielding, (b) to prevent a central crack from growing to %a length of 2a. %Given: w = 50 mm, P=150 kN, n = 3, and a=10 mm. %w=0.05; P=50*10^3; n=3; a=0.01; w=50; P=150*10^3; n=3; a=10; %Assumption: The plate will be made of Ti-6AI-6V alloy. %Solution Through the use of Table 4.2, we have K=66*sqrt(1000)sqrt(mm) and sigma_yp = 1149 MPa K_c=66*sqrt(1000); sigma_yp = 1149*10^6; %a. The permissible tensile stress on the basis of the net area is %sigma_all=sigma_yp/n=P/(2*(w-a)*t t=P*n/(2*(w-a)*(10^-3)*sigma_yp)

%b. From the case A of Table 4.2, a_div_w=a/w lambda=1.03 %Applying Eq. (4.18), the stress at fracture is sigma=K_c/(lambda*n*sqrt(pi*a)) sigma_all=sigma; %Inasmuch as this stress is smaller than the yield strength, the fracture governs the design; %sigma_all=120.5 MPa %Hence, t_reg=P/(2*w*sigma_all) %Comment Use a thickness of 13 mm. Both values of a and t satisfy Table 4.3. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 4.6: t= 0.0049 a_div_w = 0.2000 lambda = 1.0300 sigma = 120.5065 t_reg = 12.4475 Results in textbook notation with units:

(a) t = 4.9 mm (b)  = 120.5 MPa, treg = 12.45 mm >> %EXAMPLE 4.8 Fatigue Pressure of a Cylindrical Tank %Consider a thin-walled cylindrical tank of radius r = 120 mm and thickness t = 5 mm, subject to an %internal pressure varying from a value of p/4 to p. r=0.12; t=0.005; %Employ the octahedral shear theory together with the Soderberg criterion to compute the value of p %producing failure after 108 cycles. The material tensile yield strength is 300 MPa and the fatigue %strength is sigma_cr =250 MPa at 10^8 cycles. ten_streng=300*10^6; fati_streng=250*10^6; 18

%Solution The maximum and minimum values of the tangential and axial principal stresses are given by sigma_t_max_div_p=r/t sigma_t_min_div_p=(-1/4)*r/t sigma_z_max_div_p=r/(2*t) sigma_z_min_div_p=(-1/4)*r/(2*t) %The alternating and mean stresses are therefore sigma_t_a=(sigma_t_max_div_p-sigma_t_min_div_p)/2 sigma_z_a=(sigma_z_max_div_p-sigma_z_min_div_p)/2 %and sigma_t_m=(sigma_t_max_div_p+sigma_t_min_div_p)/2 sigma_z_m=(sigma_z_max_div_p+sigma_z_min_div_p)/2 %The octahedral shearing stress theory, Eq. (4.6), for cyclic combined stress is expressed as sigma_e_a=sqrt(sigma_t_a^2-(sigma_t_a)*(sigma_z_a)+sigma_z_a^2) sigma_e_m=sqrt(sigma_t_m^2-(sigma_t_m)*(sigma_z_m)+sigma_z_m^2) %from which sigma_e_a = 12.99p and sigma_e_m = 7.794p. The Soderberg relation then leads to p=1/((sigma_e_a/fati_streng)+(sigma_e_m/ten_streng)) %Solving this equation, p = 12.82 MPa ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 4.8 : sigma_t_max_div_p = 24 sigma_t_min_div_p = -6 sigma_z_max_div_p = 12 sigma_z_min_div_p = -3 sigma_t_a = 15 sigma_z_a = 7.5000 sigma_t_m = 9 sigma_z_m =

4.5000 sigma_e_a = 12.9904 sigma_e_m = 7.7942 p= 1.2830e+07 Results in textbook notation with units:

 ea = 12.99 p,  em = 7.794 p, p = 12.83 MPa >> %EXAMPLE 4.10 Dynamic Stress and Deflection of a Metal Beam %A weight is dropped from a height striking at midspan a simply supported beam of length The beam %is of rectangular cross section: a = 25 mm width and b = 75 mm depth. For a material with modulus of %elasticity W=180; h=0.1; L=1.16; a=0.025; b=0.075; E=200*10^9; %determine the instantaneous maximum deflection and maximum stress for the following cases: (a) the %beam is rigidly supported (Fig. 4.20); (b) the beam is supported at each end by springs of stiffness k=180*10^3; %Solution The deflection of a point at midspan, owing to a statically applied load, is I=a*b^3/12.; delta_st=W*L^3/(48*E*I) %The maximum static stress, also occurring at midspan, is calculated from M=W*L; c=b/2.; sigma_st_max=M*c/(4*I) %a. The impact factor is, from Eq. (4.28), K=1+sqrt(1+(2*h/delta_st)) %We thus have delta_max=K*delta_st sigma_max=K*sigma_st_max %b. The static deflection of the beam due to its own bending and the deformation of the spring is delta_st_sp=delta_st+(W/2)/k %The impact factor is thus K_sp=1+sqrt(1+(2*h/delta_st_sp)) %Hence, delta_max=K_sp*delta_st_sp sigma_max=K_sp*sigma_st_max

%Comments It is observed from a comparison of the results that dynamic loading increases the value of %deflection and stress considerably. Also noted is a reduction in stress with increased flexibility %attributable to the springs added to the supports. The values calculated for the dynamic stress are %probably somewhat high, because h >> delta_st in both cases. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 4.10: delta_st = 3.3299e-05 sigma_st_max = 2227200 K= 78.5059 delta_max = 0.0026 sigma_max = 1.7485e+08 delta_st_sp = 5.3330e-04 K_sp = 20.3913 delta_max = 0.0109 sigma_max = 4.5416e+07 Results in textbook notation with units:

 st = 0.033 10−3 m,  st ,max = 2.23 MPa (a) K = 78.5,  max = 26 mm,  max = 174.9 MPa (b)  st = 0.533 10−3 mm, K = 20.4,  max = 10.87 mm,  max = 45.49 MPa >> %EXAMPLE 5.1 Analysis of an Angle in Pure Bending

%A 150- by 150-mm slender angle of 20-mm thickness is subjected to oppositely directed end couples %at the centroid of the cross section. %What bending stresses exist at points A and B on a section away from the ends (Fig. 5.4a)? %Determine the orientation of the neutral axis. %Solution Equations (5.13) and (5.16) are applied to ascertain the normal stress. This requires first the %determination of a number of section properties through the use of familiar expressions of mechanics %given in Appendix C. y1=130; z1=20; zb1=10; z2=150; y2=20; zb2=75; A1=130*20; A2=150*20; M_z=11*1000; zb=(y1*z1*zb1+z2*y2*zb2)/(A1+A2) %For a rectangular section of depth h and width b, the moment of inertia about the neutral axis is %I_z=b*h^3/12 (Table C.1). using the parallel-axis theorem. Applying Eq. (C.9), d_y1=40; d_z1=-35; d_y2=-35; d_z2=30; I_z=z1*y1*y1*y1/12+y1*z1*d_y1*d_y1+z2*y2*y2*y2/12+z2*y2*d_y2*d_y2 I_y=I_z %The transfer formula (C.11) for a product of inertia yields I_yz_b1=0.; I_yz_b2=0.; I_yz=I_yz_b1+A1*d_y1*d_z1+I_yz_b2+A2*d_y2*d_z2 %Stresses Using Formula (5.13). We have y_A=0.105m, y_B=-0.045m, z_A=-0.045m, z_B=-0.045m, %M_y=0 y_A=0.105; y_B=-0.045; z_A=-0.045; z_B=-0.045; M_y=0; I_z=I_z/1000000; I_y=I_y/1000000; I_yz=I_yz/1000000; sigma_x_A=M_z*(I_yz*z_A-I_y*y_A)/(I_y*I_z-I_yz*I_yz) sigma_x_B=M_z*(I_yz*z_B-I_y*y_B)/(I_y*I_z-I_yz*I_yz) %Alternatively, these stresses may be calculated by proceeding as follows. %Directions of the Principal Axes and the Principal Moments of Inertia.Employing Eq. (5.18), we have theta_p1=(1/2.)*atan((-2*I_yz)/(I_y-I_z)); theta_p1_d=theta_p1*180/pi theta_p2_d=theta_p1_d+90 %Therefore, the two values of theta_p are 45 and 135 Substituting the first of these values into Eq. (5.17), we obtain . I_y_p=(I_y+I_z)/2.+((I_y-I_z)/2.)*cos(2*theta_p1)-I_yz*sin(2*theta_p1); %Since the principal moments of inertia are, by application of Eq. (5.19), I_1=(I_y+I_z)/2.+sqrt(((I_z-I_y)/2.)*((I_z-I_y)/2.)+I_yz*I_yz) I_2=(I_y+I_z)/2.-sqrt(((I_z-I_y)/2.)*((I_z-I_y)/2.)+I_yz*I_yz) I_z_p=I_2; %it is observed that I_1=I_y_p=18.386*10^6 mm^4 and I_2=I_z_p=4.806*10^6 mm^4 %The principal axes are indicated in Fig. 5.4b as the y_p_z_p axes. %Stresses Using Formula (5.16). The components of bending moment about the principal axes are M_y_p=M_z*sin(theta_p1) M_z_p=M_z*cos(theta_p1) %Equation (5.16) is now applied, referring to Fig. 5.4b, with y_A_p=0.043 m, z_A_p=-0.106 m, y_B_p=%0.063 m and z_B_p=0 determined from geometrical considerations: y_A_p=0.043; z_A_p=-0.106; y_B_p=-0.0636; z_B_p=0; 22

sigma_x_A=(M_y_p*z_A_p/I_y_p)-(M_z_p*y_A_p/I_z_p) sigma_x_B=(M_y_p*z_B_p/I_y_p)-(M_z_p*y_B_p/I_z_p) %as before Direction of the Neutral Axis.From Eq. (5.15), with M_y=0, M_y=0.; phi=atan((M_y*I_z+M_z*I_yz)/(M_y*I_yz+M_z*I_y)) phi_d=phi*180/pi %The negative sign indicates that the neutral is located counterclockwise %from the z axis (Fig. 5.4b) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 5.1: zb = 44.8214 I_z = 1.1597e+07 I_y = 1.1597e+07 I_yz = -6790000 sigma_x_A = -113.5241 sigma_x_B = 102.9820 theta_p1_d = 45 theta_p2_d = 135 I_1 = 18.3867 I_2 = 4.8067 M_y_p =

7.7782e+03 M_z_p = 7.7782e+03 sigma_x_A = -114.4244 sigma_x_B = 102.9179 phi = -0.5297 phi_d = -30.3495 Results in textbook notation with units:

z = 45 mm, I z = I y = 11.597 106 mm 4 , I yz = −6.79 106 mm 4 , ( x ) A = −114 MPa, ( x ) B = 103 MPa,  p = 45o , 135o ,  = −30.45o >> %EXAMPLE 5.3

Stresses in a Beam of T-Shaped Cross Section

%A simplly supported beam of length L carries a concentrated load P (Fig. 5.10a). %Find: (a) The maximum shear stress, the shear flow q_j , and the shear stress tau_j in the joint between %the flange and the web; (b) the maximum bending stress. %Given: P = 5 kN and L = 4 m. P=5*10^3; L=4; %Solution The distance from Z axis to the centroid is obtained as (Fig. 5.10a), y1=20; x1=60; A1=y1*x1; yb1=70; y2=60; x2=20; A2=y2*x2; yb2=30; y_b=(A1*yb1+A2*yb2)/(A1+A2) %The moment of inertia I about the NA is determined by the parallel axis theorem: d_y1=20; d_y2=20; I_x1=x1*y1*y1*y1/12+A1*d_y1*d_y1; I_x2=x2*y2*y2*y2/12+A2*d_y2*d_y2; I=I_x1+I_x2 %a. The maximum shearing stress in the beam takes place at the NA on the cross section supporting the %largest shear force V. %Consequently, Q_NA=y_b*x2*(y_b/2)*10^-9 I_x2=x2*y2*y2*y2/12+A2*d_y2*d_y2; %The shear force equals 2.5 kN on all cross sections of the beam (Figure 5.10b), Thus, 24

V_max=2.5*10^3 I=I*10^-12; b=x2/10^3; tau_max=V_max*Q_NA/(I*b) %The first moment of the area of the flange about the NA is y1_b=20.; Q_f=x1*y1*y1_b Q_f=Q_f*10^-9; %Shear flow and shear stress In the joint are q_j=V_max*Q_f/I tau_j=q_j/b %b. The largest moment takes place at midspan (Fig. 5.10c. Equation (5.39) is therefore M=5*10^3; y_b=y_b/10^3; c=y_b; sigma_max=M*c/I ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 5.3: y_b = 50 I= 1360000 Q_NA = 2.5000e-05 V_max = 2500 Q_f = 24000 tau_max = 2.2978e+06 q_j = 4.4118e+04 tau_j = 2.2059e+06 25

sigma_max = 1.8382e+08 Results in textbook notation with units:

y = 50 mm, I = 136 104 mm4 (a) QNA = 25 103 mm3 ,  max = 2.3 MPa, q j = 44.1 kN m,

 j = 2.206 MPa (b)  max = 183.8 MPa >> %EXAMPLE 5.7 Aluminum-Reinforced Wood Beam %A wood beam E_w=8.75 GPa 100-mm wide by 220-mm deep, has an aluminum plate E_a=70 with a %net section 80 mm by 20 mm securely fastened to its bottom face, as shown in Fig. 5.16a. Dimensions %are given in millimeters. The beam is subjected to a bending moment of around a horizontal axis. E_w=8.75*10^9; E_a=70*10^9; bw=100; hw=220; ba=80; ha=20; M=20*10^3; %Calculate the maximum stresses in both materials (a) using a transformed section of wood and (b) %using a transformed section of aluminum. %Solution %a. The modular ratio n=E_a/E_w=8. The centroid and the moment of inertia about the neutral axis of %the transformed section (Fig. 5.16b) are n=E_a/E_w; ba_t=ba*n; Aw=bw*hw; yw=hw/2; Aa=ba_t*ha; ya=hw+ha/2; y_b=(Aw*yw+Aa*ya)/(Aw+Aa) d_yw=y_b-yw; d_ya=y_b-ya; I_t=bw*hw^3/12+Aw*d_yw^2+ba_t*ha^3/12+Aa*d_ya^2 %stresses in the wood and aluminum portions are therefore c=y_b*10^-3; sigma_w_max=M*c/(I_t*10^-12) c=(hw+ha-y_b)*10^-3; sigma_a_max=n*M*c/(I_t*10^-12) %is noted that at the juncture of the two parts: c=(hw-y_b)*10^-3; sigma_w_min=M*c/(I_t*10^-12) sigma_a_min=n* sigma_w_min %b. For this case, the modular ratio n=E_w/E_a=1/8 and the transformed area is shown in Fig. 5.16c. We %now have n=E_w/E_a; bw_t=n*bw; Aw=bw_t*hw; Aa=ba*ha; I_t=bw_t*hw^3/12+Aw*d_yw^2+ba*ha^3/12+Aa*d_ya^2 %Then c=(hw+ha-y_b)*10^-3;

sigma_a_max=M*c/(I_t*10^-12) c=y_b*10^-3; sigma_w_max=n*M*c/(I_t*10^-12) %as have already been found in part (a) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 5.7: y_b = 154.1379 I_t = 2.0568e+08 sigma_w_max = 1.4988e+07 sigma_a_max = 6.6791e+07 sigma_w_min = 6.4042e+06 sigma_a_min = 5.1234e+07 I_t = 2.5711e+07 sigma_a_max = 6.6791e+07 sigma_w_max = 1.4988e+07 Results in textbook notation with units:

(a) y = 154.1 mm, I t = 205.7 106 mm 4 ,  w,max = 15 MPa,  a ,max = 66.8 MPa,

 w,min = 6.4 MPa,  a ,min = 51.2 MPa (b) I t = 25.7 106 mm 4 ,  a ,max = 66.8 MPa,  w,max = 15 MPa >>

%EXAMPLE 5.13 Maximum Stress in a Curved Rectangular Bar %A rectangular aluminum bar having mean radius r_b carries end moments M, as illustrated in Fig. 5.26. %Calculate the stresses in the member: (a) Using the flexure formula; (b) by the curved beam formula. %Given: M=1.2 kN?m, b=30 mm, h=50 mm, and r_b=125 mm. M=1.2*10^3; b=30; h=50; r_b=125; %Solution The subscripts i and o refer to the quantities of the inside and outside fibers, respectively. %a. Applying the flexure formula, Eq. (5.38) with y=h/2, we obtain y=h*10^-3/2; I=(b*10^-3)*(h*10^-3)^3/12; sigma_o=M*y/I %which is the result we would get for a straight beam. %b. We first derive the expression for the radius R of the neutral axis. From Fig. 5.26: A=b*h and %dA=bdr Integration of Eq. (5.68) between the limits r_i and r_o results %R=h/ln(r_o/r_i) %The given data leads to A=b*h r_i=r_b-h/2 r_o=r_b+h/2 %Then, Eqs. (5.73) and (5.69) yield, respectively, R=h/log(r_o/r_i) e=r_b-R %It is important to note that the radius of the neutral axis R must be calculated with five significant %figures. %The maximum compressive and tensile stresses are calculated through the use of Eq. (5.70) as %follows: A=A*10^-6; r_i=r_i*10^-3; r_o=r_o*10^-3; R=R*10^-3; e=e*10^-3; sigma_i=-M*(R-r_i)/(A*e*r_i) sigma_o=-M*(R-r_o)/(A*e*r_o) %The negative sign means a compressive stress. %Comment The maximum stress 96 MPa obtained in part (a) by the flexure formula represents an error %of about 13% from the more accurate value for the maximum stress 110.7 found in part (b). ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 5.13: sigma_o =

9.6000e+07 A= 1500 r_i = 100 r_o = 150 R= 123.3152 e= 1.6848 sigma_i = -1.1071e+08 sigma_o = 8.4471e+07 Results in textbook notation with units:

(a)  o = − i = 96 MPa (b)  i = −110.7 MPa,  o = 84.5 MPa >> %CASE STUDY 5.1 Stresses in a Steel Crane Hook by Various Methods %A load P is applied to the simple steel hook having a rectangular cross section, as illustrated in Fig. %5.27a. %Find, the tangential stresses at points A and B, using: (a) the curved beam formula; (b) the flexure %formula; (c) elasticity theory. Given: P = 6 kN, r_b = 50 mm, b = 25 mm, and h = 32 mm. P=6*10^3; r_b=50; b=25; h=32; %Solution % a. Curved Beam Formula. For the given numerical values, we obtain (Fig. 5.27b): A=b*h r_i=r_b-h/2 r_o=r_b+h/2 %Then, Eqs. (5.73) and (5.69) result in R=h/(log(r_o/r_i)) e=r_b-R

%In order to maintain applied force P in equilibrium, there must be an axial tensile force P and a %moment M=-P*r at the centroid of the section (Fig. 5.27c). Thus, by Eq. (5.74),te stress at the inner %edge (r=r_j) of the section A[nd]B: sigma_theta_A=(P/A)*( 1+(r_b*(R-r_i)/(e*r_i)) ) %(5.75a) % Likewise, the stress at the outer edge (r=r_o) sigma_theta_A=(P/A)*( 1+(r_b*(R-r_o)/(e*r_o)) ) %(5.75b) %The negative sign of sigma_theta_B means a compressive stress. The maximum tensile stress is at A %and equals 97 MPa. %Comment??The stress due to the axial force, P_div_A=P/(A*10^-6) %which is negligibly small compared to the combined stresses at points A and B of the cross section. %b. Flexure Formula. Equation (5.5), with M=-P*r gives b=b*10^-3; h=h*10^-3; r_b=r_b*10^-3; I=b*h^3/12; y=h/2; M=P*r_b sigma_theta_A=-M*y/I sigma_theta_B=-sigma_theta_A; %c. Elasticity Theory. Using Eq. (5.66) with a=r_i=34 mm and b=r we find r_o=66 mm we fine a=r_i; b=r_o; N=(1-(a/b)^2)^2-4*((a/b)^2)*(log(b/a))^2 %Superposition of [nd]P/A and the second of Eqs. (5.67) with t = 25 mm at r = a leads to t=0.025; r=a; sigma_theta_A=-P/(A*10^-6)+(4*M/(t*(b*10^-3)^2*N))*( (1-(a/b)^2)*(1+log(r/a))(1+(a/r)^2)*log(b/a) ) %Similarly, at r=b, we find r=b; sigma_theta_B=-P/(A*10^-6)+(4*M/(t*(b*10^-3)^2*N))*( (1-(a/b)^2)*(1+log(r/a))(1+(a/r)^2)*log(b/a) ) %Comments The preceding indicates that the results of the curved beam formula and elasticity theory %are in good agreement. But the flexure formula provides a result of unacceptable accuracy for the %tangential stress in this nonslender curved beam. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Case Study 5.1: A= 800 r_i = 34 r_o = 66 R= 30

48.2441 e= 1.7559 sigma_theta_A = 96.9693 sigma_theta_A = -49.9539 P_div_A = 7.5000e+06 M= 300 sigma_theta_A = -7.0312e+07 N= 0.0726 sigma_theta_A = -9.7303e+07 sigma_theta_B = 5.0536e+07 Results in textbook notation with units:

(a) (  ) A = 97 MPa, (  ) B = −50 MPa (b) (  ) B = −(  ) A = 70.3 MPa (c) (  ) A = −97.3 MPa, (  ) B = 50.5 MPa >> %EXAMPLE 6.1 Stress and Deformation in an Aluminum Shaft %A hollow aluminum alloy 6061-T6 shaft of outer radius c = 40 mm, inner radius b = 30 mm, and length %L = 1.2 m is fixed at one end and subjected to a torque T at the other end, as shown in Fig. 6.3. If the shearing stress is limited to tau_max= 140 MPa, c=0.04; b=0.03; L=1.2; tau_max=140*10^6; %find: (a) The largest value of the torque; (b) the corresponding minimum value of shear stress; (c) the %angle of twist that will create a shear stress tau_min= 100 MPa on the inner surface. tau_min=100*10^6; 31

%Solution From Table D.1, we have the shear modulus of elasticity G = 72 GPa and tau_yp= 220 MPa. G=72*10^9; tau_yp=220*10^6; %a. Inasmuch as tau_max < tau_yp, we can apply Eq. (6.1) with r = c %By Table C.1, the polar moment of inertia of the hollow circular tube is J=pi*(c^4-b^4)/2 T=J*tau_max/c %b. The smallest value of the shear stress takes place on the inner surface of the shaft, and tau_min %and tau_max are respectively proportional to b and c. Therefore, tau_min=b*tau_max/c %c. Through the use of Eq. (2.27), the shear strain on the inner surface of the shaft is equal to gamma_min=tau_min/G; gamma_min_mu=gamma_min*10^6 %Referring to Fig. 6.2, phi=L*gamma_min/b phi=phi*180/pi ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 6.1: J= 2.7489e-06 T= 9.6211e+03 tau_min = 105000000 gamma_min_mu = 1.4583e+03 phi = 0.0583 phi = 3.3423 Results in textbook notation with units:

(a) J = 2.749(106 ) mm 4 , T = 9.62 kN  m (b)  min = 105 MPa (c)  min = 1458 , rad = 0.0583 rad, deg = 3.34o >> %Example 6.7 Analysis of a Stepped Bar in Torsion %A rectangular bar of width b consists of two segments: one with depth and length and the other %with depth and length (Fig. 6.15). The bar is to be designed using an allowable shearing stress and %an allowable angle of twist per unit length per meter. a1=0.06; L1=2.5; a2=0.045; L2=1.5; tau_all=50*10^6; theta_all=1.5; %Determine (a) the maximum permissible applied torque tau_max assuming b=30 mm and G=80GPa %and (b) the corresponding angle of twist between the end sections, phi_max b=0.03; G=80*10^9; %Solution The values of the torsion parameters from Table 6.2 are %For segment AC(a1/b=2); alpha_1=0.2406, beta_1=0.229 %For segment CB(a2/b=1.5); alpha_2=0.231, beta_2=0.196 %a. Segment BC governs because it is of smaller depth. The permissible torque T based on the allowable %shearing stress is obtained from tau_max=tau_all tau_max=tau_all; %Thus, T=alpha_2*a2*b*b*tau_max %The allowable torque T corresponding to the allowable angle of twist per unit length is determined %from theta=T/beta*a*b^3*G. T=beta_2*a2*b^3*G*theta_all*pi/180 %maximum permissible torque, equal to the smaller of the two preceding values, is T_max=468 N∙m T_max=468; %b. The angle of twist is equal to the sum of the angles of twist for the two segments: phi_max=(T_max/(b^3*G))*(L1/(beta_1*a1)+L2/(beta_2*a2)) phi_d_max=phi_max*180/pi %Clearly, had the allowable torque been based on the angle of twist per unit length, we would have %found that phi_max=theta_all*(L1+L2) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 6.7: alpha_1 = 0.2406 beta_1 =

0.2290 alpha_2 = 0.2310 beta_2 = 0.1960 T= 467.7750 T= 498.7592 phi_max = 0.0763 phi_d_max = 4.3700 phi_max = 6 Results in textbook notation with units:

(a) T from  max = 468 N  m, T from = 499 N  m, Tmax = 468 N  m (b) (max ) sumof theangles = 4.37o , (max )unit length = 6o >> %EXAMPLE 6.8 Rectangular Torsion Tube %An aluminum tube of rectangular cross section (Fig. 6.19a) is subjected to a torque of 56.5 kN∙m along %its longitudinal axis. T=56.5*10^3; t1=0.012; t2=0.006; t3=0.01; t4=0.006; %Determine the shearing stresses and the angle of twist. Assume G=28*10^9; %Solution Referring to Fig. 6.19b, which shows the membrane surface mnnm (representing Phi), the %applied torque is, according to Eq. (b), wd=0.5; ht=0.25; A=wd*ht; h=T/(2*A) %from which h=226,000 N/m. The shearing stresses are found from Eq. (a) as follows: tau_1=h/t1 tau_2=h/t2 tau_3=h/t3 34

tau_4=tau_2 theta=h/(2*A*G)*(ht/t1+2*wd/t2+ht/t3) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 6.8: h= 226000 tau_1 = 1.8833e+07 tau_2 = 3.7667e+07 tau_3 = 22600000 tau_4 = 3.7667e+07 theta = 0.0069 Results in textbook notation with units:

 = 0.0069 rad m >> %EXAMPLE 6.10 Analysis of an I-Beam Under Torsion %A cantilever I-beam with the idealized cross section shown in Fig. 6.22 is subjected to a torque of 1.2 %kN∙m . T=1.2*10^3; %Determine (a) the maximum longitudinal stress, and (b) the total angle of twist, phi. Take G=80 GPa %and E=200 GPa. Let t_f=10 mm t_w=7 mm, b=0.1 m, h=0.2 and L=2.4 m. G=80*10^9; E=200*10^9; t_f=0.01; t_w=0.007; b=0.1; h=0.2; L=2.4; %Solution %a. The torsional rigidity of the beam is, from Eq. (6.21a), b1=h-t_f; b2=b; t1=t_w; t2=t_f; C_div_G=(b1*t1^3+2*b2*t2^3)/3 %The flexural rigidity of one flange is I_f=t2*b2^3/12 35

C=G*C_div_G %Hence, from Eq. (6.28) we have one_div_alpha_div_h=sqrt(E*I_f/(2*C)) M_f= T*one_div_alpha_div_h; %From Eq. (6.30), the bending moment in the flange is found to be 3.43 times larger than the applied %torque, T. Thus, the maximum longitudinal bending stress in the flange is sigma_f_max=M_f*(b2/2)/I_f %b. Since exp(-alpha*L)=0.03, we can apply Eq. (6.31) to calculate the angle of twist at the free end: phi_free=(T/C)*(L-one_div_alpha_div_h*h) phi_fixed=(T*L/C) phi_fixed_div_phi_free=phi_fixed/phi_free %Comment It is interesting to note that if the ends of the beam were both free, the total angle of twist %would be phi=(T*L/C)=0.4073 rad, and the beam would experience phi_free/phi_fixed=1.4 times more %twist under the same torque. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 6.10: C_div_G = 8.8390e-08 I_f = 8.3333e-07 C= 7.0712e+03 one_div_alpha_div_h = 3.4329 sigma_f_max = 2.4717e+08 phi_free = 0.2908 phi_fixed = 0.4073 phi_fixed_div_phi_free = 1.4007

Results in textbook notation with units:

(a)  f ,max = 247 MPa (b)  freeend = 0.2908 rad >> %EXAMPLE 7.6 Properties of a Truss Bar Element %The element 1-2 of the steel truss shown in Fig. 7.13 with a length L, cross-sectional area A, and %modulus of elasticity E, is oriented at angle counterclockwise from the x axis. Given:u1=0.5 mm, %v1=0.625 mm, u2=-1.25, v2=0, theta=30o , A=600 mm2 , L=1.5 m, E=200 GPa u1=0.5; v1=0.625; u2=-1.25; v2=0; theta=30.; A=600*10^-6; L=1.5; E=200*10^9; %Find: (a) the global stiffness matrix for the element; (b) the local displacements u1_bar, v1_bar. %u2_bar, and v2_bar of the element; (c) the axial stress in the element. %Solution The free-body diagram of the element 1- 2 is shown in Fig. 7.15. The spring rate of the %element is AE_div_L=A*E/L %and rd=30*pi/180; c=cos(rd), s=sin(rd) %a. Element Stiffness Matrix. Applying Eq. (7.38), k1=(A*E/L)*[ c^2 c*s -c^2 -c*s; c*s s^2 -c*s -s^2; -c^2 -c*s c^2 c*s; -c*s -s^2 c*s s^2 ]; k1_div_AE_L=[ c^2 c*s -c^2 -c*s; c*s s^2 -c*s -s^2; -c^2 -c*s c^2 c*s; -c*s -s^2 c*s s^2 ] %b. Element Displacements. Equations (7.35b), {del_bar}_e=[T]{del}_e results in del_e=[u1 v1 u2 v2] tdel_e=transpose(del_e); T=[ c s 0 0; -s c 0 0; 0 0 c s; 0 0 -s c ] del_bar_e=T*tdel_e %c. Axial Force. Substituting the given numerical values into Eq. (7.39) with i = 1 and j = 2, we find k_f12=(A*E/L)*[c s] del_f12=[u2-u1 v2-v1]; %get del_f12 transpose matrix tdel_f12=transpose(del_f12) F_12=k_f12*tdel_f12*10^-3 F_1=F_12; 37

sigma_1=F_1/A %It follows that axial stress in the element is equal to sigma_1=F_1/A = -244 MPa. %Comment The negative sign indicates a compression ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 7.6: AE_div_L = 8.0000e+07 c= 0.8660 s= 0.5000 k1_div_AE_L = 0.7500 0.4330 -0.7500 -0.4330 0.4330 0.2500 -0.4330 -0.2500 -0.7500 -0.4330 0.7500 0.4330 -0.4330 -0.2500 0.4330 0.2500 del_e = 0.5000 0.6250 -1.2500

T= 0.8660 0.5000

-0.5000 0.8660

0 0.8660 0.5000

0 -0.5000 0.8660

del_bar_e = 0.7455 0.2913 -1.0825 0.6250 k_f12 =

1.0e+07 * 6.9282 4.0000 tdel_f12 = -1.7500 -0.6250 F_12 = -1.4624e+05 sigma_1 = -2.4367e+08 Results in textbook notation with units:

 0.7500 0.4330 -0.7500 -0.4330   0.4330 0.2500 -0.4330 -0.2500  6  N m (a ) K = 80(10 )  -0.7500 -0.4330 0.7500 0.4330     -0.4330 -0.2500 0.4330 0.2500   u1   0.7455     0.2913      (b)  1  =   mm u2   -1.0825   0.6250   2    (c) F1 = F12 = −146.2 kN,  1 = −244 MPa >> %CASE STUDY 7.1 Analysis of a Three-Bar Truss %A steel plane truss in which all members have the same axial rigidity AE supports a horizontal force P %and a load W acting at joint 2, as shown in Fig. 7.17a. %Find the nodal displacements, reactions, and stresses in each member. Given: P=24 kN, W=36 kN, %sigma_yp=250 MPa, E=200 GPa, L_1=L_2=L=2 m, L_3=2*sqrt(2), A=400 mm2 P=24*10^3; W=36*10^3; sigma_yp=250*10^6; E=200*10^9; L_1=2; L_2=L_1; L=L_2; L_3=2*sqrt(2); A=400*10^-6; %Input Data. At each node there are two displacements and two nodal force components (Fig. 7.17b). It %is recalled that theta is measured counterclockwise from the positive x axis to each element (Table %7.2). %Element 1 Stiffness Matrix. Applying Eq. (7.38), theta=0.*pi/180; c=cos(theta); s=sin(theta); k1=4*10^7*[ c^2 c*s -c^2 -c*s; c*s s^2 -c*s -s^2; -c^2 -c*s c^2 c*s; 39

-c*s -s^2 c*s s^2 ] %Element 2 Stiffness Matrix. Applying Eq. (7.38), theta=270.*pi/180; c=cos(theta); s=sin(theta); k2=4*10^7*[ c^2 c*s -c^2 -c*s; c*s s^2 -c*s -s^2; -c^2 -c*s c^2 c*s; -c*s -s^2 c*s s^2 ] %Element 3 Stiffness Matrix. Applying Eq. (7.38), theta=225.*pi/180; c=cos(theta); s=sin(theta); k3=2*sqrt(2)*10^7*[ c^2 c*s -c^2 -c*s; c*s s^2 -c*s -s^2; -c^2 -c*s c^2 c*s; -c*s -s^2 c*s s^2 ] %To make 6x6 matrix k6_1=zeros(6); k6_2=zeros(6); k6_3=zeros(6); K=zeros(6); %Adding each element for k1, k2, and k3 to k6_1, k6_2, and k6_3 k6_1(1,1)=k1(1,1); k6_1(1,2)=k1(1,2); k6_1(1,3)=k1(1,3); k6_1(1,4)=k1(1,4); k6_1(2,1)=k1(2,1); k6_1(2,2)=k1(2,2); k6_1(2,3)=k1(2,3); k6_1(2,4)=k1(2,4); k6_1(3,1)=k1(3,1); k6_1(3,2)=k1(3,2); k6_1(3,3)=k1(3,3); k6_1(3,4)=k1(3,4); k6_1(4,1)=k1(4,1); k6_1(4,2)=k1(4,2); k6_1(4,3)=k1(4,3); k6_1(4,4)=k1(4,4); k6_1 k6_2(1,1)=k2(1,1); k6_2(1,2)=k2(1,2); k6_2(1,5)=k2(1,3); k6_2(1,6)=k2(1,4); k6_2(2,1)=k2(2,1); k6_2(2,2)=k2(2,2); k6_2(2,5)=k2(2,3); k6_2(2,6)=k2(2,4); k6_2(5,1)=k2(3,1); k6_2(5,2)=k2(3,2); k6_2(5,5)=k2(3,3); k6_2(5,6)=k2(3,4); k6_2(6,1)=k2(4,1); k6_2(6,2)=k2(4,2); k6_2(6,5)=k2(4,3); k6_2(6,6)=k2(4,4); k6_2 k6_3(3,3)=k3(1,1); k6_3(3,4)=k3(1,2); k6_3(3,5)=k3(1,3); k6_3(3,6)=k3(1,4); k6_3(4,3)=k3(2,1); k6_3(4,4)=k3(2,2); k6_3(4,5)=k3(2,3); k6_3(4,6)=k3(2,4); k6_3(5,3)=k3(3,1); k6_3(5,4)=k3(3,2); k6_3(5,5)=k3(3,3); k6_3(5,6)=k3(3,4); k6_3(6,3)=k3(4,1); k6_3(6,4)=k3(4,2); k6_3(6,5)=k3(4,3); k6_3(6,6)=k3(4,4); k6_3 K=k6_1+k6_2+k6_3 sel_K=zeros(3); sel_K(1,1)=K(3,3); sel_K(1,2)=K(3,4); sel_K(1,3)=K(3,6); sel_K(2,1)=K(4,3); sel_K(2,2)=K(4,4); sel_K(2,3)=K(4,6); 40

sel_K(3,1)=K(6,3); sel_K(3,2)=K(6,4); sel_K(3,3)=K(6,6); sel_K inv_sel_K=inv(sel_K) F=[P -W 0] tF=transpose(F); d=inv_sel_K*tF RK(1,1)=K(1,3);RK(1,2)=K(1,4);RK(1,3)=K(1,6); RK(2,1)=K(2,3);RK(2,2)=K(2,4);RK(2,3)=K(2,6); RK(3,1)=K(5,3);RK(3,2)=K(5,4);RK(3,3)=K(5,6); RK R=RK*d %The results may be verified by applying the equilibrium equations to the free-body diagram of the %entire truss, Fig. 7.17a. %Axial Forces in Elements. From Eqs. (7.39) and (d) and Table 7.2, we obtain c=1; s=0; u_2=d(1); v_2=d(2); m=[c s]; disp=[u_2 v_2]; tdisp=transpose(disp); F12=(A*E/L)*m*tdisp c=1; s=-1; u_3=0; v_3=d(3); m=[c s]; disp=[u_3 v_3]; tdisp=transpose(disp); F13=(A*E/L)*m*tdisp c=-1/sqrt(2); s=-1/sqrt(2); u_2=d(1); v_2=d(2); v_3=d(3); m=[c s]; disp=[-u_2 v_3-v_2]; tdisp=transpose(disp); L=sqrt(2)*L; F23=(A*E/(sqrt(2)*L))*m*tdisp %Stresses in Elements. By dividing the preceding element forces by the cross-sectional area of each bar, %we obtain sigma_1=F12/A sigma_2=sigma_1*F13/F12 sigma_3=sigma_1*F23/F12 %The negative sign indicates a compressive stress. %Comment The results indicate that member axial stresses are well below the yield strength for the %material considered. Observe that the FEA permits the calculation of displacements, forces, and %stresses in the truss with unprecedented ease and precision. It is evident, however, that the FEA, even %in the simplest cases, requires considerable algebra. For any significant problem, the electronic digital %computer must be used. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Case Study 7.1: k1 = 40000000 0 -40000000

0 -40000000 0

0 40000000

0 41

k2 = 1.0e+07 * 0.0000 0.0000 -0.0000 -0.0000 0.0000 4.0000 -0.0000 -4.0000 -0.0000 -0.0000 0.0000 0.0000 -0.0000 -4.0000 0.0000 4.0000 k3 = 1.0e+07 * 1.4142 1.4142 -1.4142 -1.4142 1.4142 1.4142 -1.4142 -1.4142 -1.4142 -1.4142 1.4142 1.4142 -1.4142 -1.4142 1.4142 1.4142 k6_1 = 40000000 0

0 -40000000 0

-40000000

0 0

0 40000000

0 0

k6_2 = 1.0e+07 * 0.0000 0.0000

0 -0.0000 -0.0000

0.0000 4.0000

0 -0.0000 -4.0000

-0.0000 -0.0000

0 0.0000 0.0000

-0.0000 -4.0000

0 0.0000 4.0000

k6_3 = 42

1.0e+07 * 0

0 1.4142 1.4142 -1.4142 -1.4142

0 -1.4142 -1.4142 1.4142 1.4142

K= 1.0e+07 * 4.0000 0.0000 -4.0000 0.0000 4.0000 -4.0000 0

0 -0.0000 -0.0000 0 -0.0000 -4.0000

0 5.4142 1.4142 -1.4142 -1.4142 0 1.4142 1.4142 -1.4142 -1.4142

-0.0000 -0.0000 -1.4142 -1.4142 1.4142 1.4142 -0.0000 -4.0000 -1.4142 -1.4142 1.4142 5.4142 sel_K = 1.0e+07 * 5.4142 1.4142 -1.4142 1.4142 1.4142 -1.4142 -1.4142 -1.4142 5.4142 inv_sel_K = 1.0e-06 * 0.0250 -0.0250

-0.0250 0.1207 0.0250 0 0.0250 0.0250 F= 24000

-36000

d= 0.0015 43

-0.0049 -0.0009 RK = 1.0e+07 * -4.0000 0

0 -0.0000 0 -4.0000

-1.4142 -1.4142 1.4142 R= 1.0e+04 * -6.0000 3.6000 3.6000 F12 = 6.0000e+04 F13 = 36000 F23 = -5.1000e+04 sigma_1 = 1.5000e+08 sigma_2 = 90000000 sigma_3 = -1.2750e+08 Results in textbook notation with units:

u2   0.0015      2  = -0.0049 m   -0.0009  3    R1x  -6.0000     4  R1y  =  3.6000  (10 )N R   3.6000   3x    F12 = 6(104 ) N, F13 = 3.6(104 ) N, F23 = −5.1(104 ) N

 1 = 150 MPa,  2 = 90 MPa,  3 = −127.5 MPa >> %EXAMPLE 8.2 Thick-Walled Cylinder Pressure Requirement %A steel cylinder is subjected to an internal pressure four times greater than the external pressure. The %tensile elastic strength of the steel is ?yp = 340MPa and the shearing elastic strength tau_yp = %sigmasigma_yp/2 = 170 MPa sigma_yp=340*10^6; tau_yp=sigma_yp/2; %Calculate the allowable internal pressure according to the various yielding theories of failure. The %dimensions are a = 0.1 and b = 0.15 Let nu = 0.3. a=0.1; b=0.15; nu=0.3; %Solution The maximum stresses occur at the innermost fibers. From Eqs. (8.8), for r = a and p_i = 4p_o %we have r=a; sigma_theta_div_p_i=(a^2-b^2*(1/4))/(b^2-a^2)+((1-(1/4))*a^2*b^2)/((b^2-a^2)*r^2) sigma_r_div_p_i=(a^2-b^2*(1/4))/(b^2-a^2)-((1-(1/4))*a^2*b^2)/((b^2-a^2)*r^2) %The value of internal pressure at which yielding begins is predicted according to the various theories of %failure, as follows: %a. Maximum shearing stress theory: p_i_Max_sh_theory=tau_yp/(( sigma_theta_div_p_i- sigma_r_div_p_i)/2) %b. Energy of distortion theory [Eq. (4.5a)]: p_i_Energy_dist_theory=sigma_yp/sqrt(sigma_theta_div_p_i^2+ sigma_r_div_p_i ^2sigma_theta_div_p_i* sigma_r_div_p_i) %c. Octahedral shearing stress theory: By use of Eqs. (4.6) and (1.38)and(1.43), we have p_i_Octa_sh_theory=(sqrt(2)*sigma_yp/3)/((sqrt((sigma_theta_div_p_i- sigma_r_div_p_i)^2+ sigma_r_div_p_i^2+(- sigma_theta_div_p_i)^2))/3) %Comments The results found in (b) and (c) are identical as expected (Sec. 4.8). The onset of inelastic %action is governed by the maximum shearing stress: the allowable value of internal pressure is limited %to 125.9 MPa, modified by an appropriate factor of safety. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 8.2:

sigma_theta_div_p_i = 1.7000 sigma_r_div_p_i = -1 p_i_Max_sh_theory = 1.2593e+08 p_i_Energy_dist_theory = 1.4380e+08 p_i_Octa_sh_theory = 1.4380e+08 Results in textbook notation with units:

(a) pi = 125.9 MPa (b) pi = 143.8 MPa (c) pi = 143.8 MPa >> %EXAMPLE 8.3 Stresses in a Compound Cylinder under Internal Pressure %A compound cylinder with a = 150 mm, b = 200 mm, c = 250 mm, E = 200 GPa, and delta= 0.1 mm is %subjected to an internal pressure of 140 MPa. a=0.15; b=0.2; c=0.25; E=200*10^9; delta=0.0001; p_i=140*10^6; %Determine the distribution of tangential stress throughout the composite wall. %Solution In the absence of applied internal pressure, the contact pressure is, from Eq. (8.23), p=(E*delta/b)*((b^2-a^2)*(c^2-b^2))/(2*b^2*(c^2-a^2)) %The tangential stresses in the outer cylinder associated with this pressure are found by using Eq. (8.13) r=b sigma_theta_b=(b^2*p/(c^2-b^2))*(1+c^2/r^2) r=c sigma_theta_c=(b^2*p/(c^2-b^2))*(1+c^2/r^2) %The stresses in the inner cylinder are, from Eq. (8.16), r=a sigma_theta_a=-(p*b^2/(b^2-a^2))*(1+a^2/r^2) r=b sigma_theta_b=-(p*b^2/(b^2-a^2))*(1+a^2/r^2) %The preceding stresses are plotted in Fig. 8.7, indicated by the dashed lines kk and mm. The stresses %owing to internal pressure alone, through the use of Eq. (8.13) with b = c are found to be %(sigma_theta)r=0.15 = 297.5 Mpa (sigma_theta)r=0.02 = 201.8 MPa, (sigma_theta)r=0.25 = 157.5 MPa, %and are shown as the dashed line nn. The stress resultant is obtained by superposition of the two %distributions, represented by the solid line.

%Comments The use of a compound prestressed cylinder has thus reduced the maximum stress from %297.5 to 257.8 MPa. Based on the maximum principal stress theory of elastic failure, significant weight %savings can apparently be effected through such configurations. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 8.3: p= 1.2305e+07 r= 0.2000 sigma_theta_b = 5.6055e+07 r= 0.2500 sigma_theta_c = 4.3750e+07 r= 0.1500 sigma_theta_a = -5.6250e+07 r= 0.2000 sigma_theta_b = -4.3945e+07 Results in textbook notation with units:

p = 12.3 MPa, outer cylinder :(  ) r =0.2 = 56.1 MPa, (  ) r =0.25 = 43.8 MPa, inner cylinder :(  )r =0.15 = −56.3 MPa, (  ) r =0.2 = 43.9 MPa >> %EXAMPLE 8.6 Rotating Shrink-Fit Performance Analysis

%A flat 0.5-m outer diameter, 0.1-m inner diameter, and 0.075-m-thick steel disk is shrunk onto a steel %shaft (Fig. 8.12). If the assembly is to run at speeds up to n = 6900 rpm, do=0.5; di=0.1; t=0.075; n=6900; %determine (a) the shrinking allowance, (b) the maximum stress when not rotating, and (c) the %maximum stress when rotating. The material properties are rho=7.8 kn∙s^2/m^4, E=200 GPa and nu = %0.3. rho=7.8; E=200*10^9; nu=0.3; %Solution %a. The radial displacements of the disk (ud) and shaft (us) are, from Eqs. (8.28) and (8.30), r=di/2; a=di/2; b=do/2; ud_div_rho_omega_E=r*((3+nu)*(1-nu)/8)*(a^2+b^2-(1+nu)*r^2/(3+nu)+((1+nu)/(1nu))*((a^2*b^2)/r^2)) us_div_rho_omega_E=r*((1-nu)/8)*((3+nu)*b^2-(1+nu)*r^2) %We observe that us may be neglected, as it is less than 1% of ud of the disk at the common radius. The %exact allowance is delta=(ud_div_rho_omega_E- us_div_rho_omega_E)*(rho*(n*2*pi/60)^2/E) %b. Applying Eq. (8.23), we have b=di/2; c=do/2; p=(E*delta/(2*b))*((c^2-b^2)/c^2) r=di/2; b=di/2; c=do/2; %Therefore, from Eq. (8.18), sigma_theta_max=p*(c^2+b^2)/(c^2-b^2) %c. From Eq. (8.28), for sigma_theta_max_div_rho_omega=((3+nu)/8)*(a^2+b^2-(1+3*nu)*r^2/(3+nu)+(a^2*b^2)/r^2) sigma_theta_max=sigma_theta_max_div_rho_omega * (rho*(n*2*pi/60)^2) %A plot of the variation of stress in the rotating disk is shown in Fig. 8.12. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 8.6: ud_div_rho_omega_E = 0.0026 us_div_rho_omega_E = 2.1875e-05 delta = 5.2496e-05 p= 1.0079e+08 sigma_theta_max = 1.0919e+08 48

sigma_theta_max_div_rho_omega = 0.052 sigma_theta_max = 2.1176e+08 Results in textbook notation with units:

(a)  = 5.25(10−5 ) m (b)   ,max = 109.2 MPa (c)   ,max = 0.052  2 = 211.76 MPa >> %EXAMPLE 8.7 Maximum Speed of a Flywheel Assembly %A flywheel of 380-mm diameter is to be shrunk onto a 60-mm diameter shaft %(Fig. 8.11). Both members are made of steel with rho=7.8 kn∙s^2/m^4 , %E=200 GPa and df=0.38; ds=0.06; rho=7.8*10^3; E=200*10^9; a=ds/2; b=df/2; %nu = 0.3. At a maximum speed of n = 6000 rpm, a contact pressure of p = 20 MPa is to be maintained. nu=0.3; n=6000; p=20*10^6; omega=n*2*pi/60; %Find: (a) The required radial interference; (b) the maximum tangential stress in the assembly; (c) the speed at which the fit loosens, that is, contact pressure becomes zero. %a. Through the use of Eq. (8.34), we obtain delta=(a*p/E)*(((a^2+b^2)/( b^2-a^2))+nu)+ (a*p/E)*(1-nu)+(a*rho*omega^2/(4*E))*((1nu)*a^2+(3+nu)*b^2) %(a) %Substituting the given numerical values p = 20 MPa and omega = 6000(2*pi/60) = 628.32 rad/s, Eq. (a) %results in delta = 0.02 mm. %b. Applying Eq. (8.32), we obtain sigma_theta_max=p*((a^2+b^2)/( b^2-a^2))+(rho*omega^2/4)*((1-nu)*a^2+(3+nu)*b^2) %c. Carrying delta = 0.02 × 10^-3 m and p = 0 into Eq. (a) leads to %constant of omega C=(a*rho/(4*E))*((1-nu)*a^2+(3+nu)*b^2); omega=sqrt(delta/C) %We thus have n = omega*(60/(2*pi)) %rpm. Note that, at this speed, the shrink fit becomes completely ineffective. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 8.7: delta = 1.9983e-05 49

sigma_theta_max = 1.1322e+08 omega = 755.2790 n= 7.2124e+03 Results in textbook notation with units:

(a)  = 0.02 mm (b)   ,max = 113.2 MPa (c)  = 755.3 rad s , n = 7212 rpm >> %EXAMPLE 9.4 Semi-Infinite Beam Loaded at Its End %A 2-m-long steel bar (E = 210 GPa) of 75-mm by 75-mm square cross section rests with a side on a %rubber foundation (k = 24 MPa). If a concentrated load P = 20 kN is applied at the left end of the beam %(Fig. 9.5), L=2; E=210*10^9; b=0.075; k=24*10^6; P=20*10^3; I=b^4/12; %determine (a) the maximum deflection and (b) the maximum bending stress. %Solution Applying Eq. (9.3), we have beta=(k/(4*E*I))^(1/4) beta_L=beta*L %Inasmuch as , the beam can be considered to be a long beam (see Sec. 9.5); Eqs. (9.12) with thus %apply. %a. The maximum deflection occurs at the %left end for which f4(beta x) is a maximum or beta x=0 . The first of Eqs. (9.12) is therefore format short e v_max=2*P*beta/k format short %b. Referring to Table 9.1, f2(beta x) has its maximum of 0.3224 at beta x=pi/4. Using the third of Eqs. %(9.12), f2=0.3224; M_max=P*f2/beta %The maximum stress in the beam is obtained from the flexure formula: c=b/2; I=b^4/12; sigma_max=M_max*c/I %The location of this stress is at x=pi/(4*beta) %x=pi/(4*beta) 433 mm from the left end. ----------------------------------------------------------------------------------------------------------------------------------------50

>> Results - Example 9.4: beta = 1.8143e+00 beta_L = 3.6287e+00 v_max = 3.0239e-03 M_max = 3.5539e+03 sigma_max = 5.0545e+07 x= 0.4329 Results in textbook notation with units:

 = 1.814 m −1 , (a)max = 3.02 mm, (b) M max = 3.55 kN  m,  max = 50.55 MPa >> %EXAMPLE 9.5 Finite-Length Beam with a Concentrated Load Supported by Springs %A series of springs, spaced so that a=1.5 m supports a long thin-walled steel tube having E=206.8 GPa . %A weight of 6.7 kN acts down at midlength of the tube. The average diameter of the tube is 0.1 m, and %the moment of inertia of its section is 6*10^-6 m^4. Take the spring constant of each support to be %K=10 kN/m. a=1.5; E=206.8*10^9; P=6.7*10^3; d_avg=0.1; I=6*10^-6; K=10*10^3; %Find: The maximum moment and the maximum deflection, assuming negligible tube weight. %Solution Applying Eqs. (9.3) and (9.16), we obtain k=K/a beta=(k/(4*E*I))^(1/4) %The spacing of the springs, a < pi/(4*beta) = 4.106 m, is correct. From Eq. (9.8), we have M=P/(4*beta); c=d_avg/2; sigma_max=M*c/I v_max=P*beta/(2*k) ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 9.5:

k= 6.6667e+03 beta = 0.1914 sigma_max = 7.2912e+07 v_max = 0.0962 Results in textbook notation with units:

 = 0.1914 m −1 ,  max = 72.91 MPa, max = 96.2 mm >> %EXAMPLE 10.4

Thick-Walled Half Ring

%A load P of 5 kN is applied to a steel curved bar as depicted in Fig. 10.6a. P=5*10^3; %Determine the vertical deflection of the free end by considering the effects of the internal normal and %shear forces in addition to the bending moment. Let E=200GPa and G=80 GPa. E=200*10^9; G=80*10^9; %Solution A free-body diagram of a portion of the bar subtended by angle theta is shown in Fig. 10.6b, %where the internal forces (N and V) and moment (M) are positive as indicated. Referring to the figure, %geometric properties of the section of the bar are b=0.01; h=0.02; I=b*h^3/12, A=b*h, R=0.05; %The form factor for shear for the rectangular section is alpha=6/5 (Table 5.2). %Integration of the foregoing results in format short e delta_v=3*pi*P*R^3/(2*E*I)+3*pi*P*R/(5*A*G)+pi*P*R/(2*A*E) %(10.11) %Comment Note that if the effects of the normal and shear forces are omitted delta_v=2.21 mm with a %resultant error in deflection of approximately 1.8%. For this curved bar, in which R/c=5, the %contribution of V and N to the displacement can thus be neglected. It is common practice to omit the %first and the third terms in Eqs. (10.6) and (10.7) when R/c>4 (Sec. 5.13). ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 10.4 : I= 6.6667e-09 A=

2.0000e-04 delta_v = 2.2482e-03 Results in textbook notation with units:

 v = 2.25 mm >> %EXAMPLE 11.1 Load-Carrying Capacity of a Wood Column %A pinned-end wood bar of width b by depth h rectangular cross section (Fig. 11.6) and length L is %subjected to an axial compressive load. %Determine (a) the slenderness ratio; (b) the allowable load, using a factor safety of n. Data: b=60 mm, %h=120 mm, L=1.8 m, n=1.4, E=12 GPa, and sigma_u=55 MPa (by Table D.1) b=60; h=120; L=1.8*10^3; n=1.4; E=12*10^9; sigma_u=55*10^6; %Solution The properties of the cross-sectional area are A = bh, Ix =hb^3/12, Iy = hb^3/12, and %r=sqrt(I/A) %a. Slenderness ratio: The smallest value of r is found when the centroidal axis is parallel to the longer %side of the rectangle. Therefore, %Substituting the given numerical value, we obtain r_y=b/sqrt(12) %(a) %Since L = Le, it follows that L_r_y=L/r_y %b. Permissible load: Applying Eq. (11.6), the Euler buckling load with r_y=r and A=b*h*10^-6 P_cr=pi^2*E*A/L_r_y^2 %Comment The largest load the column can support equals, then, P = 79/1.4 = 56.4kN. Observe that, %based on material strength, the allowable load equals P_all=sigma_u*A/n %This, compared with 56.4 kN, shows the importance of buckling ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 11.1 : r_y = 1.7321e+01 L_r_y = 1.0392e+02 A= 7.2000e-03 53

P_cr = 7.8957e+04 P_all = 2.8286e+05 Results in textbook notation with units:

(a) L / ry = 103.9 (b) Pcr = 79 kN, Pall = 283 kN >> %EXAMPLE 11.3 Buckling of the Boom of a Crane %The boom of a crane, shown in Fig. 11.10, is constructed of steel, E=210 GPa ; the yield point stress is %250 MPa. The cross section is rectangular with a depth of 100 mm and a thickness of 50 mm. E=210*10^9; d=0.1; t=0.05; L=2.75; %Determine the buckling load of the column. %Solution The moments of inertia of the section are I_z=t*d^3/12; I_y=d*t^3/12; A=d*t; r=sqrt(I_y/A) L_div_r=L/r P_div_W=1/tan(15*pi/180) %The least radius of gyration is thus r=sqrt(I_y/A)=14 mm and the slenderness ratio is L/r = 194. The %Euler formula is applicable in this range. From statics, the axial force in terms of W is P = W/tan 15° = %3.732W. Applying the formula for a hinged-end column, Eq. (11.5), for buckling in the yx plane, we %have P_cr=pi^2*E*I_z/L^2 W=P_cr/P_div_W %To calculate the load required for buckling in the xz plane, we must take note of the fact that the line %of action of the compressive force passes through the joint and thus causes no moment about the y %axis at the fixed end. Therefore, Eq. (11.5) may again be applied: P_cr=pi^2*E*I_y/L^2 W=P_cr/P_div_W W_div_A=W/A %The member will thus fail by lateral buckling when the load W exceeds 76.3 kN. Note that the critical %stress Pcr/A = 76.3/0.05 = 15.26 MPa. This, compared with the yield strength of 250 MPa, indicates the %importance of buckling analysis in predicting the safe working load. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 11.3: r= 1.4434e-02 L_div_r = 54

1.9053e+02 P_div_W = 3.7321e+00 P_cr = 1.1419e+06 W= 3.0598e+05 P_cr = 2.8548e+05 W= 7.6495e+04 W_div_A = 1.5299e+07 Results in textbook notation with units:

xy - plane :W = 305.98 kN, xz - plane :W = 76.5 kN buckling load :76.5 kN. >> %EXAMPLE 12.1 Three-Bar Structure %Determine the maximum allowable plastic stress and strain in the pin-jointed structure sustaining a %vertical load P, shown in Fig. 12.4. Assume that alpha=45 and that each element is constructed of an %aluminum alloy with the following properties: sigma_yp=350 MPa, K=840 MPa, n=0.2, %A_AD=A_CD=10*10^-5 m^2, A_BD=15*10^-5 m^2, h=3 m alpha=45*pi/180; sigma_yp=350*10^6; K=840*10^6; n=0.2; A_AD=10*10^-5; A_CD=A_AD; A_BD=15*10^-5; h=3; %Solution The structure is elastically statically indeterminate, and the solution may readily be obtained %on applying Castigliano’s theorem (Sec. 10.7). Plastic yielding begins upon loading: P=sigma_yp*A_BD+2*sigma_yp*A_AD*cos(alpha) %On applying Eqs. (e), the maximum allowable stress sigma=K*n^n %occurs at the following axial and transverse strains: epsilon_1=0.2; epsilon_2=-0.1; epsilon_3=-0.1; L=h/cos(alpha) delta_BD=h*epsilon_1 delta_AD=(h/cos(alpha))*epsilon_1

%We have L = h/cos(alpha). The total elongations for instability of the bars are thus delta_BD = 3(0.2) = %0.6 m and delta_AB=delta_CD = (3/cos 45°)(0.2) = 0.85 m. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 12.1: P= 1.0200e+05 sigma = 6.0881e+08 L= 4.2426e+00 delta_BD = 6.0000e-01 delta_AD = 8.4853e-01 Results in textbook notation with units:

P = 102, 000 N,  = 608.8 MPa, 1 = n = 0.2,  2 =  3 = −0.1,

 BD = 0.6 m,  AD =  CD = 0.85 m >> %EXAMPLE 12.4 Residual Stresses in an Assembly %Figure 12.7a shows a steel bar of 750-mm^2 cross-sectional area placed between two aluminum bars, %each of 500-mm^2 cross-sectional area. The ends of the bars are attached to a rigid support on one %side and a rigid thick plate on the other. Given: E_s= 210 GPa, sigma_s_yp = 240 MPa, E_a= 70 GPa, %and sigma_a_yp = 320 MPa. %Assumption: The material is elastic[nd]plastic. A_s=750*10^-6; A_a=500*10^-6; E_s=210*10^9; sigma_s_yp=240*10^6; E_a=70*10^9; sigma_a_yp=320*10^6; %Calculate the residual stress, for the case in which applied load P is increased from zero to P_u and %removed. format short %Solution Material Behavior. At ultimate load P_u both materials yield. Either material yielding by itself %will not result in failure because the other material is still in the elastic range. We therefore have P_a=A_a*sigma_a_yp P_s=A_s*sigma_s_yp %Hence, P_u=2*P_a+P_s 56

%Applying an equal and opposite load of this amount, equivalent to a release load (Fig. 12.6b), causes %each bar to rebound elastically. %Geometry of Deformation. Condition of geometric fit, delta_a=delta_s gives L_a=360; L_s=375; P_s_p_div_P_a_p= (L_a/(A_a*E_a))/ (L_s/(A_s*E_s)) %(a) %Condition of Equilibrium. From the free-body diagram of Fig. 12.6b, %P_s_p=500-2*P_a_p % (b) %Solving Eqs. (a) and (b) we obtain P_a_p=500/(P_s_p_div_P_a_p+2); P_s_p=500-2*P_a_p; P_a_p=P_a_p*10^3 P_s_p=P_s_p*10^3 %Superposition of the initial forces at ultimate load P_u and the elastic rebound forces owing to release %of P_u results in: P_a_res=P_a_p-P_a P_s_res=P_s_p-P_s %The associated residual stresses are thus sigma_a_res=P_a_res/A_a sigma_s_res=P_s_res/A_s %Comment We note that after this prestressing process, the assembly remains elastic as long as the %value of P_u = 500 kN is not exceeded. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 12.4: P_a = 160000 P_s = 180000 P_u = 500000 P_s_p_div_P_a_p = 4.3200 P_a_p = 7.9114e+04 P_s_p = 3.4177e+05

P_a_res = -8.0886e+04 P_s_res = 1.6177e+05 sigma_a_res = -1.6177e+08 sigma_s_res = 2.1570e+08 Results in textbook notation with units:

( Pa ) res = −80.9 kN, ( Ps ) res = 161.7 kN, ( a ) res = −162 MPa, ( s ) res = 216 MPa >> %EXAMPLE 12.7 Residual Stresses in a Rectangular Beam %Figure 12.16 shows an elastoplastic beam of rectangular cross section 40 mm by 100 mm carrying a %bending moment of M. %Determine (a) the thickness of the elastic core; (b) the residual stresses following removal of the %bending moment. Given: b = 40 mm, h = 50 mm, M = 21 kN · m, sigma_yp = 240 MPa, and E = 200 GPa. b=0.04; h=0.05; M=21*10^3; sigma_yp=240*10^6; E=200*10^9; %Solution %a. Through the use of Eq. (12.9a), we have M_yp=2*b*h^2*sigma_yp/3 %Then Eq. (12.10b) leads to e= sqrt((1-(M/M_yp)*2/3)*3)*h two_e=2*e %Elastic core depicted as shaded in Fig. 12.16. %b. The stress distribution corresponding to moment M=21 kN?m is illustrated in Fig. 12.17a. The %release of moment M produces elastic stresses, and the flexure formula applies (Fig. 12.17b). Equation %(5.5) is therefore c=h; I=b*(2*h)^3/12; sigma_max_p=M*c/I %By the superimposition of the two stress distributions, we can find the residual stresses (Fig. 12.17c). %Observe that both tensile and compressive residual stresses remain in the beam. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 12.7: M_yp = 58

1.6000e+04 e= 0.0306 two_e = 0.0612 sigma_max_p = 3.1500e+08 Results in textbook notation with units: ' (a) e = 31 mm (b)  max = 315 MPa

>> %EXAMPLE 12.12 Residual Stress in a Shaft %Figure 12.25 shows a solid circular steel shaft of diameter d and length L carrying a torque T. %Determine; (a) the radius of the elastic core; (b) the angle of twist of the shaft; (c) the residual stresses %and the residual rotation when the shaft is unloaded. Assumption: The steel is taken to be an %elastoplastic material. %Given: d = 60 mm, L = 1.4 m,T=7.75 kN m, tau_yp=145 MPa, and G = 80 GPa. d=0.06; c=d/2; L=1.4; T=7.75*10^3; tau_yp=145*10^6; G=80*10^9; %Solution We have c = 30 mm and J=pi*c^4/2 %a. Radius of Elastic Core. The yield torque, applying Eq. (12.18), equals T_yp=J*tau_yp/c %Equation (12.19), substituting the values of T and tau_yp, gives rho_0=c*( 4-(3*T/T_yp) )^(1/3) %Solving,. The elastic-plastic stress distribution in the loaded shaft is illustrated in Fig. 12.26a. %b. Yield Twist Angle. Through the use of Eq. (6.3), the angle of twist at the onset of yielding, phi_yp=T_yp*L/(G*J) %Introducing the value found for phi_yp into Eq. (12.22), we have C=c; phi=C*phi_yp/rho_0 phi_d=phi*180/pi %c. Residual Stresses and Rotation. The removal of the torque produces elastic stresses as depicted in %Fig. 12.26b, and the torsion formula, Eq. (6.1), leads to reversed stress as tau_max_p=T*c/J %Superposition of the two distributions of stress results in the residual stresses (Fig. 12.26c). %Permanent Twist. The elastic rebound rotation, using Eq. (6.3), equals phi_p=T*L/(G*J) phi_p_d=phi_p*180/pi 59

%The preceding results indicate that residual rotation of the shaft is phi_res=phi_d-phi_p_d %Comment We see that even though the reversed stresses tau_max_p exceed the yield strength %tau_yp, the assumption of linear distribution of these stresses is valid, inasmuch as they do not exceed %2tau_yp. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 12.12: J= 1.2723e-06 T_yp = 6.1497e+03 rho_0 = 0.0181 phi_yp = 0.0846 phi = 0.1403 phi_d = 8.0363 tau_max_p = 1.8273e+08 phi_p = 0.1066 phi_p_d = 6.1074 phi_res = 1.9289 Results in textbook notation with units:

(a ) o = 18.1 mm, (b) rad = 0.1403 rad, deg = 8.0363o ' ' (c) rad = 0.1066 rad, deg = 6.1074o , res = 1.93o

>> %EXAMPLE 13.5 Compressed Air Tank %A steel cylindrical vessel with hemispherical ends or so-called heads, supported by two cradles (Fig. %13.16), contains air at a pressure of p. %Calculate: (a) The stresses in the tank if each portion has the same mean radius r and the thickness t; %(b) radial extension of the cylinder; (c) change in length and thickness of the cylinder. %Given: r = 0.5 m, t = 10 mm, L = 2 m, p = 1.5 MPa, E = 200 GPa, and nu = 0.3. r=0.5; t=0.01; L=2; p=1.5*10^6; E=200*10^9; nu=0.3; %Assumptions: One of the cradles is designed so that it does not exert any axial force on the vessel: the %cradles act as simple supports. The weight of the tank may be disregarded. %Solution %a. The axial stress in the cylinder and the tangential stresses in the spherical heads are the same. %Referring to Eqs. (13.51), we write sigma_a=p*r/(2*t) sigma=sigma_a; %The tangential stress in the cylinder is thus sigma_theta=p*r/t %Comment The largest radial stress, occurring on the surface of the tank, sigma_r=p, is negligibly small %compared with the tangential and axial stresses. The state of stress in the wall of a thin-walled vessel is %thus taken to be biaxial. %b. Through the use of Eq. (13.52), increase in radius is delta_c=(p*r^2/(2*E*t))*(2-nu) %c. Applying the generalized Hoooke’s law, we have sigma_r=0; epsilon_a=(1/E)*(sigma_a-nu*(sigma_r+sigma_theta)) epsilon_theta=(1/E)*(sigma_theta-nu*(sigma_r+sigma_a)) epsilon_r=(1/E)*(sigma_r-nu*(sigma_theta+sigma_a)) del_L=epsilon_a*L del_t=epsilon_r*t %Since Delta_L = epsilon_a*L, and Delta_t = epsilon_r*t, we obtain Delta_L =75*(10^-3)*2 = 0.15 mm, %Delta_t = -168*(10^-3)*(0.01) = -0.0012 mm. %Comment: Observe that there are small increase in the length and negligibly small decrease in the %wall thickness of the vessel. ---------------------------------------------------------------------------------------------------------------------------------------->> Results - Example 13.5: sigma_a =

37500000 sigma_theta = 75000000 delta_c = 1.5938e-04 epsilon_a = 7.5000e-05 epsilon_theta = 3.1875e-04 epsilon_r = -1.6875e-04 del_L = 0.15 del_t = -0.0017 Results in textbook notation with units:

(a)  a = 37.5 MPa,   = 75 MPa (b)  c = 0.16 mm (c) L = 0.15 mm, t = −0.0017 mm >> End of MATLAB Solutions